Argent 2015

Problema 1: a.

Sistema de coordenada en la posición del barco enemigo con $x_M=2500$ m posición de la montaña, $y_M=1800$ m altura de la montaña y $x_E=2800$ m posición de la costa este de la isla. $\theta =45^{\circ }$ es el ángulo de disparo.

$a_x=0$ $a_y=-g=-10\text{m}{\text{s}}^{-2}$

$v_x(t)=v_0\cos \theta =\frac{\sqrt{2}}{2}{v}_0=0.7071v_0$ $v_y(t)=v_0\mathrm{sen}\theta -gt=\frac{\sqrt{2}}{2}{v}_0-10\text{m}{\text{s}}^{-2}t=0.7071v_0-10\text{m}{\text{s}}^{-2}t$

$x(t)=v_0\cos \theta t=\frac{\sqrt{2}}{2}{v}_{0}t=0.7071v_{0}t$ $y(t)=v_0\mathrm{sen}\theta t-\frac{1}{2}g{t}^2=\frac{\sqrt{2}}{2}{v}_{0}t-5\text{m}{\text{s}}^{-2}{t}^2=0.7071v_{0}t-5\text{m}{\text{s}}^{-2}{t}^2$

b.

Como $x=v_0\cos (\theta )t$ entonces,

$t=\frac{x}{v_0\cos (\theta )}$

Reemplazando $t$ en $y(t)$

$y(x)=\tan \theta x-\frac{g{x}^2}{2v_0^2\cos (\theta {)}^2}=x-\frac{10\text{m}{\text{s}}^2}{v_0^2}{x}^2$

c. Dado que $v_0=720$ km ${\text{h}}^{-1}$ = 200 m ${\text{s}}^{-1}$ entonces,

$y(x)=x-2.5\times 10^{-4}{\text{m}}^{-2}{x}^2$

Para que el proyectil supere la montaña es necesario que la trayectoria cumpla que,

$y_M>y(x_M)$

Como $y(x_M)=937.5$ m $<1800$ m $=y_M$ el proyectil no supera la montaña y choca a una altura (respecto del nivel del agua) de 937.5 m.

d. La velocidad mínima $v_{\text{min}}$ necesaria para que el proyectil es aquella en la cual la trayectoria cumple,

$y_M=x_M-\frac{g}{v_{\text{min}}^2}{x}_M^2$

Despejando se obtiene que

$v_{\text{min}}=\sqrt{\frac{g{x}_M^2}{x_M-y_M}}=298.807\text{m}{\text{s}}^{-1}$

e. La distancia más cercana a la orilla este a la cual un proyectil puede impactar está dado por las raíces de la trayectoria obtenida utilizando la velocidad mínima determinada en el punto anterior,

$y(x)=x-\frac{g}{v_{\text{min}}^2}{x}^2=x\left(1-\frac{g}{v_{\text{min}}^2}x\right)=x(1-1.12\times 10^{-4}x)$

Las raíces $x_1$ y $x_2$ son

$x_1=0\text{m}$ $x_2=8928.571\text{m}$

$x_1$ corresponde a la posición del barco enemigo y $x_2$ a la distancia a la cual el proyectil impacta en el agua. Entonces el rango de distancia medida desde la orilla este es,

$[0;6128.571]\text{m}$

barco, isola, montagna 1800m, traiettoria

sistema di coordinate per proiettile

Topic: Newtonian Mechanics Metodi: Kinematic Equations, Vector Decomposition, Physical Modeling Competenze: Mathematical Modeling, Physical Reasoning Objects: Projectile Fonte: Testo (PDF) — p.15

Problema 1: a.

Sistema di coordinate in posizione del navetto nemico con $x_M=2500$ m posizione della nave montagna, $y_M=1800$ m altezza della montagna e $x_E=2800$ m posizione della costa orientale dell’isola. $\theta =45^{\circ }$ è l’angolo di lancio.

$a_x=0$ $a_y=-g=-10\text{m}{\text{s}}^{-2}$

$v_x(t)=v_0\cos \theta =\frac{\sqrt{2}}{2}{v}_0=0.7071v_0$ $v_y(t)=v_0\mathrm{sen}\theta -gt=\frac{\sqrt{2}}{2}{v}_0-10\text{m}{\text{s}}^{-2}t=0.7071v_0-10\text{m}{\text{s}}^{-2}t$

$x(t)=v_0\cos \theta t=\frac{\sqrt{2}}{2}{v}_{0}t=0.7071v_{0}t$ $y(t)=v_0\mathrm{sen}\theta t-\frac{1}{2}g{t}^2=\frac{\sqrt{2}}{2}{v}_{0}t-5\text{m}{\text{s}}^{-2}{t}^2=0.7071v_{0}t-5\text{m}{\text{s}}^{-2}{t}^2$

b.

Como $x=v_0\cos (\theta )t$ entonces,

$t=\frac{x}{v_0\cos (\theta )}$

Sostituendo $t$ da $y(t)$

$y(x)=\tan \theta x-\frac{g{x}^2}{2v_0^2\cos (\theta {)}^2}=x-\frac{10\text{m}{\text{s}}^2}{v_0^2}{x}^2$

c. Poiché $v_0=720$ km ${\text{h}}^{-1}$ = 200 m ${\text{s}}^{-1}$ allora,

$y(x)=x-2.5\times 10^{-4}{\text{m}}^{-2}{x}^2$

Per superare la montagna, il proiettile deve essere in orario che:

$y_M>y(x_M)$

Come $y(x_M)=937.5$ m $<1800$ m $=y_M$ il proiettile non supera la montagna e colpisce una altezza (rispetto al livello dell’acqua) di 937,5 m.

d. La velocità minima $v_{\text{min}}$ necessaria per il proiettile è quella in cui il proiettile percorso complesso,

$y_M=x_M-\frac{g}{v_{\text{min}}^2}{x}_M^2$

La chiusura ottiene che

$v_{\text{min}}=\sqrt{\frac{g{x}_M^2}{x_M-y_M}}=298.807\text{m}{\text{s}}^{-1}$

e. La distanza più vicina alla costa orientale che un proiettile può colpire è data per le radici del percorso ottenuto utilizzando la velocità minima determinata nel punto precedente,

$y(x)=x-\frac{g}{v_{\text{min}}^2}{x}^2=x\left(1-\frac{g}{v_{\text{min}}^2}x\right)=x(1-1.12\times 10^{-4}x)$

Le radici $x_1$ e $x_2$ sono:

$x_1=0\text{m}$ $x_2=8928.571\text{m}$

$x_1$ corrisponde alla posizione della nave nemico e $x_2$ alla distanza a cui il proiettile colpisce l’acqua. Quindi il range di distanza misurato dalla riva orientale è,

$[0;6128.571]\text{m}$

barco, isola, montagna 1800m, traiettoria

sistema di coordinate per proiettile

Topic: Newtonian Mechanics Metodi: Kinematic Equations, Vector Decomposition, Physical Modeling Competenze: Mathematical Modeling, Physical Reasoning Objects: Projectile Fonte: Testo (PDF) — p.15

Problem number one: a.

Coordinate system in enemy ship position with $x_M=2500$ m position mountain, $y_M=1800$ m mountain height and $x_E=2800$ m position of the east coast of the island. $\theta =45^{\circ }$ is the angle of shooting.

$a_x=0$ $a_y=-g=-10\text{m}{\text{s}}^{-2}$

$v_x(t)=v_0\cos \theta =\frac{\sqrt{2}}{2}{v}_0=0.7071v_0$ $v_y(t)=v_0\mathrm{sen}\theta -gt=\frac{\sqrt{2}}{2}{v}_0-10\text{m}{\text{s}}^{-2}t=0.7071v_0-10\text{m}{\text{s}}^{-2}t$

$x(t)=v_0\cos \theta t=\frac{\sqrt{2}}{2}{v}_{0}t=0.7071v_{0}t$ $y(t)=v_0\mathrm{sen}\theta t-\frac{1}{2}g{t}^2=\frac{\sqrt{2}}{2}{v}_{0}t-5\text{m}{\text{s}}^{-2}{t}^2=0.7071v_{0}t-5\text{m}{\text{s}}^{-2}{t}^2$

b.

Como $x=v_0\cos (\theta )t$ entonces,

$t=\frac{x}{v_0\cos (\theta )}$

Substituting $t$ for $y(t)$

$y(x)=\tan \theta x-\frac{g{x}^2}{2v_0^2\cos (\theta {)}^2}=x-\frac{10\text{m}{\text{s}}^2}{v_0^2}{x}^2$

c. Since $v_0=720$ km ${\text{h}}^{-1}$ = 200 m ${\text{s}}^{-1}$ then,

$y(x)=x-2.5\times 10^{-4}{\text{m}}^{-2}{x}^2$

For the projectile to overcome the mountain, the trajectory must meet the following criteria:

$y_M>y(x_M)$

As $y(x_M)=937.5$ m $<1800$ m $=y_M$ the projectile does not overrun the mountain and hits a height (in respect of water level) of 937.5 m.

d. The minimum speed $v_{\text{min}}$ required for the projectile is that at which the projectile is trajectory is fulfilled,

$y_M=x_M-\frac{g}{v_{\text{min}}^2}{x}_M^2$

Clearing you get that

$v_{\text{min}}=\sqrt{\frac{g{x}_M^2}{x_M-y_M}}=298.807\text{m}{\text{s}}^{-1}$

e. The closest distance to the east bank to which a projectile can strike is given The roots of the trajectory obtained using the minimum speed determined at the previous point,

$y(x)=x-\frac{g}{v_{\text{min}}^2}{x}^2=x\left(1-\frac{g}{v_{\text{min}}^2}x\right)=x(1-1.12\times 10^{-4}x)$

The roots $x_1$ and $x_2$ are

$x_1=0\text{m}$ $x_2=8928.571\text{m}$

$x_1$ corresponds to the position of the enemy ship and $x_2$ to the distance at which the projectile is It impacts the water. So the range of distance measured from the east bank is,

$[0;6128.571]\text{m}$

barco, isola, montagna 1800m, traiettoria

sistema di coordinate per proiettile

Topic: Newtonian Mechanics Metodi: Kinematic Equations, Vector Decomposition, Physical Modeling Competenze: Mathematical Modeling, Physical Reasoning Objects: Projectile Fonte: Testo (PDF) — p.15

Problema 2: a.

b. Cuerpo 1.

${\vec{T}}_1+{\vec{P}}_1=m_1{\vec{a}}_1$

Según el sistema de coordenada elegido para el cuerpo 1

$T_1-m_{1}g=-m_{1}a$

Cuerpo 2.

${\vec{T}}_2+{\vec{P}}_2+{\vec{F}}_R+\vec{N}=m_2{\vec{a}}_2$

Según el sistema de coordenada elegido para el cuerpo 2 En la dirección x) $T_2-F_R-m_{2}g\sin 30=m_{2}a$ En la dirección y) $N-m_{2}g\cos 30=0$

Como la cuerda no tiene masa y no hay fricción en las poleas,

$T_1=T_2=T$

c. De la ecuación de movimiento del cuerpo 1

$T=m_1(g-a)=48\text{N}$

d. De la ecuación de movimiento en la dirección x) del cuerpo 2

$F_R=T-m_2(a+g\sin 30)=20\text{N}$

e. Como $F_R={\mu }_{d}N$

De la ecuación de movimiento en la dirección y) del cuerpo 2 $N=m_{2}g\cos 30$

Luego

${\mu }_d=\frac{F_R}{N}=\frac{T-m_2(a+g\sin 30)}{m_{2}g\cos 30}=\frac{1}{\sqrt{3}}$

blocchi collegati da fune su piano inclinato 30°

diagramma corpo libero Cuerpo 1 (soluzione)

diagramma corpo libero Cuerpo 2 (soluzione)

diagramma corpo libero entrambi i blocchi

Topic: Newtonian Mechanics Metodi: Free-Body Diagram, Vector Decomposition, Physical Modeling Competenze: Diagrammatic Reasoning, Mathematical Modeling, Physical Reasoning Objects: Block, String, Pulley, Inclined Plane Fonte: Testo (PDF) — p.17

Il problema 2: a.

b. Corpo uno.

${\vec{T}}_1+{\vec{P}}_1=m_1{\vec{a}}_1$

Secondo il sistema di coordinate scelto per il corpo 1

$T_1-m_{1}g=-m_{1}a$

Corpo 2.

${\vec{T}}_2+{\vec{P}}_2+{\vec{F}}_R+\vec{N}=m_2{\vec{a}}_2$

Secondo il sistema di coordinate scelto per il corpo 2 In direzione x) $T_2-F_R-m_{2}g\sin 30=m_{2}a$ In indirizzo e) $N-m_{2}g\cos 30=0$

Poiché la corda non ha massa e non c’è attrito nelle pollice,

$T_1=T_2=T$

c. Della equazione di movimento del corpo 1

$T=m_1(g-a)=48\text{N}$

d. dell’equazione di movimento nella direzione x) del corpo 2

$F_R=T-m_2(a+g\sin 30)=20\text{N}$

e. Come $F_R={\mu }_{d}N$

dell’equazione di movimento in direzione y) del corpo 2 $N=m_{2}g\cos 30$

Poi

${\mu }_d=\frac{F_R}{N}=\frac{T-m_2(a+g\sin 30)}{m_{2}g\cos 30}=\frac{1}{\sqrt{3}}$

blocchi collegati da fune il suo pianoforte inclinato 30°

*diagrafo corpo libero Corpo 1 (soluzione) *

*diagrafo corpo libero Corpo 2 (soluzione) *

diagrafo corpo libero entrambi i blocchi

Topic: Newtonian Mechanics Metodi: Free-Body Diagram, Vector Decomposition, Physical Modeling Competenze: Diagrammatic Reasoning, Mathematical Modeling, Physical Reasoning Objects: Block, String, Pulley, Inclined Plane Fonte: Testo (PDF) — p.17

Problem two: a.

b. Body one.

${\vec{T}}_1+{\vec{P}}_1=m_1{\vec{a}}_1$

According to the coordinate system chosen for the body 1

$T_1-m_{1}g=-m_{1}a$

Body two.

${\vec{T}}_2+{\vec{P}}_2+{\vec{F}}_R+\vec{N}=m_2{\vec{a}}_2$

According to the coordinate system chosen for body 2 In the address x) $T_2-F_R-m_{2}g\sin 30=m_{2}a$ In the address y) $N-m_{2}g\cos 30=0$

Since the rope has no mass and no friction in the pulleys,

$T_1=T_2=T$

c. From the equation of motion of the body 1

$T=m_1(g-a)=48\text{N}$

d. Of the equation of motion in the direction x) of the body 2

$F_R=T-m_2(a+g\sin 30)=20\text{N}$

e. As $F_R={\mu }_{d}N$

Of the equation of motion in the direction y) of the body 2 $N=m_{2}g\cos 30$

Then

${\mu }_d=\frac{F_R}{N}=\frac{T-m_2(a+g\sin 30)}{m_{2}g\cos 30}=\frac{1}{\sqrt{3}}$

The following is the list of the instruments used in the production of the instrument:

The following table shows the results of the evaluation of the results of the evaluation:

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The Commission has also adopted a draft decision on the implementation of the Community’s budget.

Topic: Newtonian Mechanics Metodi: Free-Body Diagram, Vector Decomposition, Physical Modeling Competenze: Diagrammatic Reasoning, Mathematical Modeling, Physical Reasoning Objects: Block, String, Pulley, Inclined Plane Fonte: Testo (PDF) — p.17

Problema 3:

a. La energía inicial del sistema es

$E_0=\frac{1}{2}m{v}^2+mg{h}_1=4\text{J}$

Por conservación de energía $E_A=\frac{1}{2}m{v}_A^2+mg{h}_2=4\text{J}$

$v_A=5\text{m}{\text{s}}^{-1}$

b. Por conservación de energía, cuando el resorte se comprimió totalmente se debe cumplir que $E_R=\frac{1}{2}{k}_1{x}^2+mg{h}_2=4\text{J}$

$x=0.2\text{m}$

c. Por conservación de energía, la velocidad del bloque en la loma tiene igual magnitud pero dirección contraria a la velocidad inicial,

$v=4\text{m}{\text{s}}^{-1}$

d. Por conservación de energía hasta el punto B

$E_B=\frac{1}{2}m{v}_B^2=4\text{J}$

$v_B=6.325\text{m}{\text{s}}^{-1}$

Entre los puntos B y C hay roce dinámico por lo cual no se conserva la energía en ese tramo. El módulo de la fuerza de roce es,

$F_R={\mu }_{d}N={\mu }_{d}mg$

La perdida de energía es,

$W=-F_{R}L=-{\mu }_{d}mgL=-1.3\text{J}$

$W=E_C-E_B=\frac{1}{2}m{v}_C^2-4\text{J}$

$v_C=5.196\text{m}{\text{s}}^{-1}$

e. La energía se conserva desde C hasta el resorte y desde el resorte a C. Al pasar por el tramo entre los puntos C y B, el bloque pierde la misma energía -1.3J que cuando recorrió el tramo la primera vez, por lo cual en el punto B la energía del bloque es,

$E_B=4\text{J}-1.3\text{J}-1.3\text{J}=1.4\text{J}$

La mínima energía para alcanzar la elevación $h_1$ es,

$E_{\text{min}}=mg{h}_1=2.4\text{J}$

Por lo tanto, el bloque no alcanza la elevación $h_1$. La altura $h$ a la cual llega el bloque es,

$E_B=mgh=1.4\text{J}$

$h=0.7\text{m}$

Prueba Experimental Hoja de respuestas.

Puntos 1 a 7. Por cada medición: 1.50 ptos.

medición	D [cm]	d [cm]	D’ [cm]	d’ [cm]	M [g]
1	1.8	12.2	4.1	9.9	18.7
2	1.6	10.4	3.5	8.5	19.4
3	1.7	11.3	3.8	9.2	19.0
4	1.5	9.5	3.3	7.7	18.5
5	1.4	8.6	3.0	7.0	18.8
6	1.2	7.8	2.7	6.3	18.2
7	1.1	6.9	2.4	5.6	18.6
8	1.0	6.0	2.1	4.9	19.1
9	0.8	5.2	1.8	4.2	18.2
10	0.7	4.3	1.5	3.5	18.8

Punto 8: 5.00 ptos. $M=\bar{M}\pm \frac{s}{\sqrt{10}}=18.7\pm 0.1\text{g}$

SOLUCION

medición	D [cm]	d [cm]	D’ [cm]	d’ [cm]	M [g]
1	1.8	12.2	4.1	9.9	18.7
2	1.6	10.4	3.5	8.5	19.4
3	1.7	11.3	3.8	9.2	19.0
4	1.5	9.5	3.3	7.7	18.5
5	1.4	8.6	3.0	7.0	18.8
6	1.2	7.8	2.7	6.3	18.2
7	1.1	6.9	2.4	5.6	18.6
8	1.0	6.0	2.1	4.9	19.1
9	0.8	5.2	1.8	4.2	18.2
10	0.7	4.3	1.5	3.5	18.8

Las distancias se miden con 1 mm de incertidumbre y la masa de agua con 0.2 g si la jeringa tiene 5 subdivisiones por cada ${\text{cm}}^3$

$\bar{M}={\sum }_i{M}_i=18.73\text{g}$

$s=\sqrt{\frac{{\sum }_i(M_i-\bar{M}{)}^2}{N}}=0.36\text{con }N=10$

$M=\bar{M}\pm \frac{s}{\sqrt{10}}=18.7\pm 0.1\text{g}$

Otra opción es asociar una incertidumbre igual al ancho del intervalo de masas determinadas dividido 2. En este caso:

$\epsilon =\frac{M_{\text{max}}-M_{\text{min}}}{2}=0.6\text{g}$

$M=18.7\pm 0.6\text{g}$

blocco su loma con due molle R1 e R2

Topic: Conservation of Energy, Newtonian Mechanics Metodi: Energy Conservation Method, Conservation of Energy, Statistical Averaging Competenze: Physical Reasoning, Experimental Data Analysis, Error Propagation Objects: Block, Spring Fonte: Testo (PDF) — p.19

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p.9 — bilancia a bracci con pesi (Figura 1)

p.9 — regola su bordo tavolo con tapita (Figura 2)

Problema 3:

a. L’energia iniziale del sistema è

$E_0=\frac{1}{2}m{v}^2+mg{h}_1=4\text{J}$

Per la conservazione dell’energia $E_A=\frac{1}{2}m{v}_A^2+mg{h}_2=4\text{J}$

$v_A=5\text{m}{\text{s}}^{-1}$

b. Per conservare l’energia, quando la risorsa è stata completamente compressa deve essere rispettata che $E_R=\frac{1}{2}{k}_1{x}^2+mg{h}_2=4\text{J}$

$x=0.2\text{m}$

c. Per conservare l’energia, la velocità del blocco nella loma è uguale ma in direzione contraria alla velocità iniziale,

$v=4\text{m}{\text{s}}^{-1}$

d. Per conservazione dell’energia fino al punto B

$E_B=\frac{1}{2}m{v}_B^2=4\text{J}$

$v_B=6.325\text{m}{\text{s}}^{-1}$

Tra i punti B e C c’è un rottura dinamico che non consente di conservare l’energia in quel punto.

• La trama. Il modulo della forza di rottura è,

$F_R={\mu }_{d}N={\mu }_{d}mg$

La perdita di energia è,

$W=-F_{R}L=-{\mu }_{d}mgL=-1.3\text{J}$

$W=E_C-E_B=\frac{1}{2}m{v}_C^2-4\text{J}$

$v_C=5.196\text{m}{\text{s}}^{-1}$

e. L’energia viene conservata da C fino alla primavera e dalla primavera fino a C. Passando per il tratto Tra i punti C e B, il blocco perde la stessa energia -1.3J che quando ha percorso il la prima volta, quindi al punto B l’energia del blocco è,

$E_B=4\text{J}-1.3\text{J}-1.3\text{J}=1.4\text{J}$

L’energia minima per raggiungere l’elevazione $h_1$ è:

$E_{\text{min}}=mg{h}_1=2.4\text{J}$

Il blocco non raggiunge quindi l’elevazione $h_1$. L’altezza $h$ a cui raggiunge il blocco è,

$E_B=mgh=1.4\text{J}$

$h=0.7\text{m}$

Prova sperimentale Pagina di risposte.

Punti da uno a sette. Per ogni misurazione: 1,50 p.

medizione d [cm] d [cm] d [cm] d [cm] d [cm] m [g] |---|---|---|---|---|---| | 1 | 1.8 | 12.2 | 4.1 | 9.9 | 18.7 | | 2 | 1.6 | 10.4 | 3.5 | 8.5 | 19.4 | | 3 | 1.7 | 11.3 | 3.8 | 9.2 | 19.0 | | 4 | 1.5 | 9.5 | 3.3 | 7.7 | 18.5 | | 5 | 1.4 | 8.6 | 3.0 | 7.0 | 18.8 | | 6 | 1.2 | 7.8 | 2.7 | 6.3 | 18.2 | | 7 | 1.1 | 6.9 | 2.4 | 5.6 | 18.6 | | 8 | 1.0 | 6.0 | 2.1 | 4.9 | 19.1 | | 9 | 0.8 | 5.2 | 1.8 | 4.2 | 18.2 | | 10 | 0.7 | 4.3 | 1.5 | 3.5 | 18.8 |

Sotto il punto: $M=\bar{M}\pm \frac{s}{\sqrt{10}}=18.7\pm 0.1\text{g}$

SOLUCIONE

medizione d [cm] d [cm] d [cm] d [cm] d [cm] m [g] |---|---|---|---|---|---| | 1 | 1.8 | 12.2 | 4.1 | 9.9 | 18.7 | | 2 | 1.6 | 10.4 | 3.5 | 8.5 | 19.4 | | 3 | 1.7 | 11.3 | 3.8 | 9.2 | 19.0 | | 4 | 1.5 | 9.5 | 3.3 | 7.7 | 18.5 | | 5 | 1.4 | 8.6 | 3.0 | 7.0 | 18.8 | | 6 | 1.2 | 7.8 | 2.7 | 6.3 | 18.2 | | 7 | 1.1 | 6.9 | 2.4 | 5.6 | 18.6 | | 8 | 1.0 | 6.0 | 2.1 | 4.9 | 19.1 | | 9 | 0.8 | 5.2 | 1.8 | 4.2 | 18.2 | | 10 | 0.7 | 4.3 | 1.5 | 3.5 | 18.8 |

Le distanze sono misurate con 1 mm di incertezza e la massa dell’acqua con 0,2 g se: la siringa ha 5 suddivisioni per ogni ${\text{cm}}^3$

$\bar{M}={\sum }_i{M}_i=18.73\text{g}$

$s=\sqrt{\frac{{\sum }_i(M_i-\bar{M}{)}^2}{N}}=0.36\text{con }N=10$

$M=\bar{M}\pm \frac{s}{\sqrt{10}}=18.7\pm 0.1\text{g}$

Un’altra opzione è associare un’incertezza uguale alla larghezza dell’intervallo di massa

• di cui al punto 2. In questo caso:

$\epsilon =\frac{M_{\text{max}}-M_{\text{min}}}{2}=0.6\text{g}$

$M=18.7\pm 0.6\text{g}$

blocco la sua loma con due molle R1 e R2

Topic: Conservation of Energy, Newtonian Mechanics Metodi: Energy Conservation Method, Conservation of Energy, Statistical Averaging Competenze: Physical Reasoning, Experimental Data Analysis, Error Propagation Objects: Block, Spring Fonte: Testo (PDF) — p.19

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p.9 bilanciamento a bracci con peso (Figura 1)

**p.9 ** regola il bordo della tavola con tappeto (Figura 2)

Problem three:

a. The initial energy of the system is

$E_0=\frac{1}{2}m{v}^2+mg{h}_1=4\text{J}$

Energy conservation $E_A=\frac{1}{2}m{v}_A^2+mg{h}_2=4\text{J}$

$v_A=5\text{m}{\text{s}}^{-1}$

b. For energy conservation, when the spring is fully compressed it must be met which $E_R=\frac{1}{2}{k}_1{x}^2+mg{h}_2=4\text{J}$

$x=0.2\text{m}$

c. For energy conservation, the block velocity in the loop is equal to the magnitude but direction contrary to the initial speed,

$v=4\text{m}{\text{s}}^{-1}$

d. By energy conservation up to point B

$E_B=\frac{1}{2}m{v}_B^2=4\text{J}$

$v_B=6.325\text{m}{\text{s}}^{-1}$

Between points B and C there is dynamic friction so no energy is conserved in that I’ll take the stretch. The modulus of bridging force is,

$F_R={\mu }_{d}N={\mu }_{d}mg$

The energy loss is,

$W=-F_{R}L=-{\mu }_{d}mgL=-1.3\text{J}$

$W=E_C-E_B=\frac{1}{2}m{v}_C^2-4\text{J}$

$v_C=5.196\text{m}{\text{s}}^{-1}$

e. Energy is conserved from C to spring and from spring to C. Passing the stretch Between points C and B, the block loses the same energy -1.3J as when it traveled the So at point B the energy of the block is,

$E_B=4\text{J}-1.3\text{J}-1.3\text{J}=1.4\text{J}$

The minimum energy to reach the $h_1$ elevation is,

$E_{\text{min}}=mg{h}_1=2.4\text{J}$

Therefore, the block does not reach the $h_1$ elevation. The height $h$ at which the block reaches is,

$E_B=mgh=1.4\text{J}$

$h=0.7\text{m}$

The test shall be carried out in accordance with the following conditions: Answer sheet.

Scores are 1-7. For each measurement: 1.50 pts.

♪ ♪ Measurement ♪ ♪ D [cm] ♪ D [cm] ♪ D [cm] ♪ M [g] ♪ |---|---|---|---|---|---| | 1 | 1.8 | 12.2 | 4.1 | 9.9 | 18.7 | | 2 | 1.6 | 10.4 | 3.5 | 8.5 | 19.4 | | 3 | 1.7 | 11.3 | 3.8 | 9.2 | 19.0 | | 4 | 1.5 | 9.5 | 3.3 | 7.7 | 18.5 | | 5 | 1.4 | 8.6 | 3.0 | 7.0 | 18.8 | | 6 | 1.2 | 7.8 | 2.7 | 6.3 | 18.2 | | 7 | 1.1 | 6.9 | 2.4 | 5.6 | 18.6 | | 8 | 1.0 | 6.0 | 2.1 | 4.9 | 19.1 | | 9 | 0.8 | 5.2 | 1.8 | 4.2 | 18.2 | | 10 | 0.7 | 4.3 | 1.5 | 3.5 | 18.8 |

The point is 8:5.00 p.m. $M=\bar{M}\pm \frac{s}{\sqrt{10}}=18.7\pm 0.1\text{g}$

The Commission

♪ ♪ Measurement ♪ ♪ D [cm] ♪ D [cm] ♪ D [cm] ♪ M [g] ♪ |---|---|---|---|---|---| | 1 | 1.8 | 12.2 | 4.1 | 9.9 | 18.7 | | 2 | 1.6 | 10.4 | 3.5 | 8.5 | 19.4 | | 3 | 1.7 | 11.3 | 3.8 | 9.2 | 19.0 | | 4 | 1.5 | 9.5 | 3.3 | 7.7 | 18.5 | | 5 | 1.4 | 8.6 | 3.0 | 7.0 | 18.8 | | 6 | 1.2 | 7.8 | 2.7 | 6.3 | 18.2 | | 7 | 1.1 | 6.9 | 2.4 | 5.6 | 18.6 | | 8 | 1.0 | 6.0 | 2.1 | 4.9 | 19.1 | | 9 | 0.8 | 5.2 | 1.8 | 4.2 | 18.2 | | 10 | 0.7 | 4.3 | 1.5 | 3.5 | 18.8 |

Distances are measured with 1 mm of uncertainty and the water mass with 0.2 g if the syringe has 5 subdivisions for each ${\text{cm}}^3$

$\bar{M}={\sum }_i{M}_i=18.73\text{g}$

$s=\sqrt{\frac{{\sum }_i(M_i-\bar{M}{)}^2}{N}}=0.36\text{con }N=10$

$M=\bar{M}\pm \frac{s}{\sqrt{10}}=18.7\pm 0.1\text{g}$

Another option is to associate an uncertainty equal to the width of the mass interval. the following: In this case:

$\epsilon =\frac{M_{\text{max}}-M_{\text{min}}}{2}=0.6\text{g}$

$M=18.7\pm 0.6\text{g}$

block its rim with two R1 and R2 springs

Topic: Conservation of Energy, Newtonian Mechanics Metodi: Energy Conservation Method, Conservation of Energy, Statistical Averaging Competenze: Physical Reasoning, Experimental Data Analysis, Error Propagation Objects: Block, Spring Fonte: Testo (PDF) — p.19

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The weight of the weighted arm balances (Figure 1)

The table edge shall be adjusted with a rug (Figure 2)