View page as slide show

Practice Exam 1 Solutions

Question 1 Solution

As halloween entertainment a 4kg pumpkin is shot. The pumpkin breaks in to 3 pieces of equal mass. The bullet is lodged in one of the pieces and this piece continues in the original direction of the bullet. Another piece flies straight up and the third piece moves angle $\theta$ below the horizontal. The velocity of the piece that flies straight up is 1ms$^{-1}$ and the velocity of the third piece moving at an angle $\theta$ below the horizontal is 3ms$^{-1}$.

A. (5 points) The bullet weighs 15g and has an initial velocity of 400 ms$^{-1}$. What was the initial momentum and kinetic energy of the bullet?

$\vec{p_{i}}=0.015\times400=6\mathrm{\,kg\,m\,s^{-1}}$ right

$KE=\frac{1}{2}mv^{2}=\frac{1}{2}\times0.015\times400^{2}=1200\mathrm{\,J}$

B. (5 points) How high does the piece of pumpkin that flew straight up go above it's original position before it turns around and starts coming back down?

$v^{2}=v_{0}^{2}-2\times9.8\times(x-x_{0})$

$2\times9.8\times h=1$

$h=\frac{1}{19.6}=0.051\mathrm{\,m}=5.1\mathrm{\,cm}$

C. (5 points) What is the value of the angle $\theta$?

From conservation of momentum in the y direction

$\frac{4}{3}\times3\times \sin\theta=\frac{4}{3}\times 1$

$\sin\theta=\frac{1}{3}$

$\theta=19.47^{o}$

D. (5 points) What is the velocity of the piece of the pumpkin with the bullet in it after the collision?

From conservation of momentum in the x direction

$(\frac{4}{3}+0.015)v_{f}-\frac{4}{3}\times3\times\cos19.47^{o}=6$

$v_{f}=\frac{6+\frac{4}{3}\times3\times\cos19.47^{o}}{\frac{4}{3}+0.015}=7.25\mathrm{\,m\,s^{-1}}$

E. (5 points) How much kinetic energy was lost in this collision?

$KE_{final}=\frac{1}{2}\times\frac{4}{3}\times1^{2}+\frac{1}{2}\times\frac{4}{3}\times3^{2}+\frac{1}{2}\times(\frac{4}{3}+0.015)\times7.25^{2}=42\mathrm{\,J}$

$KE_{lost}=1158\mathrm{\,J}$

Question 2 Solution

In an overly elaborate halloween trick a large fake bat (mass 2kg) is to be raised using a rope which runs over a pulley and is wound around a wheel which can then be turned with a handle. The wheel has mass 4kg and has a radius of 20cm, while the pulley has a mass of 0.5kg and a radius of 10cm. The handle has a length of 30cm and is attached to the center of the wheel. In this problem all turning objects can be considered to have frictionless bearings and the rope has no mass. Both the pulley and the wheel can be considered to be solid disks. You may neglect the moment of inertia of the handle.

A. (5 points) Initially the system is held at rest by a force exerted on the handle. What is the magnitude of this force?

Torques on big wheel must balance, so

$F\times0.3=2\times9.8\times0.2$

$F=\frac{2\times9.8\times0.2}{0.3}=13.1\mathrm{\,N}$

In order to have the bat rise up with an acceleration of 5ms$^{-2}$, a perpendicular force is exerted on the handle.

B. (5 points) What is the tension in the rope attached to the bat while it is being raised up with an acceleration of 5ms$^{-2}$?

Consider Newton's second law applied to the bat

$T_{1}-m_{B}g=m_{B}5\mathrm{\,ms^{-2}}$

$T_{1}=2\times(9.8+5)=29.6\mathrm{\,N}$

C. (5 points) What is the tension in the rope on the other side of the pulley while the bat is being raised up with an acceleration of 5ms$^{-2}$?

Consider the sum of the torques around the pulley

$T_{2}r_{p}-T_{1}r_{p}=\frac{1}{2}m_{p}r_{p}^{2}\alpha$

$T_{2}-T_{1}=\frac{1}{2}m_{p}a$

$T_{2}=\frac{1}{2}m_{p}a+T_{1}$

$T_{2}=30.85\mathrm{\,N}$

D. (5 points) In order to have the bat rise up with an acceleration of 5ms$^{-2}$, how much force must be applied to the handle (you may assume that the force is always directed perpendicular to the handle)?

Consider the sum of the torques around the wheel

$F\times0.3-30.85\times0.2=\frac{1}{2}\times4\times0.2^2\times\frac{5}{0.2}$

$F=27.23\mathrm{N}$

E. (5 points) What is the speed of the bat when it has been raised by 1.5m?

$v^{2}=2a(x-x_{0})$

$v^{2}=2\times5\times1.5$

$v=3.87\mathrm{\,m\,s^{-1}}$

F. (10 points) How much work does the person turning the wheel have to do to raise the bat by 1.5m? (Don't forget to include kinetic energy, the bat is moving with the velocity you found in part C when it has been raised by 1.5m)

$W=\Delta KE+\delta PE=\frac{1}{2}\times2\times3.87^{2}+\frac{1}{2}\times0.5\times0.5\times0.1^{2}\times(\frac{3.87}{0.1})^2+\frac{1}{2}\times0.5\times4\times0.2^{2}\times(\frac{3.87}{0.2})^2+2\times9.81\times1.5=61.25\mathrm{\,J}$

Question 3 Solution

A skeleton is suspended by 3 ropes as shown. You can consider each arm to be a rigid uniform rod of mass 0.8kg and length 70cm. The rest of the skeleton weighs 4kg and it's center of mass is 0.6m directly below the rope which is attached to skull.

A. (10 points) What is the tension in one of the ropes attached to the end of an arm. (Hint: Consider the forces and torques on the arm exerted by the rope and the shoulder.

From balance of torques on arm calculated around shoulder

$T\sin40^{o}=0.8\sin40^{o}\times g\times 0.35$

$T\frac{0.35\times0.6\times0.8\times g}{0.7}=0.4g=3.92\mathrm{\,N}$

B. (10 points) What are the horizontal and vertical components of the force exerted by the shoulder on the arm? Specify the directions of these forces (i.e.. up, down, left, right) .

Horizontal: O N

Vertical, shoulder force needs to balance 0.8g down and 0.4 g up, so the force should be 0.4g, or 3.92N up.

C. (10 points) What is the tension in the rope attached to the skull of the skeleton?

Tension should be equal and opposite to weight of main part of body and reaction force of the arm on the shoulder joint

$T=4\times9.8+2\times3.92=47.04\mathrm{\,N}$

D. (10 points) If one of the ropes attached to an arm is cut and the arm swings down, how fast is the tip of the hand moving when the arm reaches the vertical position. You may consider the shoulder to remain in in it's original position. (Hint: Use the principle of conservation of mechanical energy)

Need the change in the height of the center of mass of the arm

$\Delta h=0.35\times\cos40^{0}-0.35=-0.0818\mathrm{\,m}$

$\Delta KE=-\Delta PE$

$\frac{1}{2}I\omega^{2}=-mg\Delta h$

$\frac{1}{2}\frac{1}{3}ml^{2}(\frac{v}{l})^2=-mg\Delta h$

$v=\sqrt{-6g\Delta h}$

$v=2.19\mathrm{\,m\,s^{-1}}$

phy131studiof15/lectures/m2p1sol.txt · Last modified: 2015/10/27 14:25 by mdawber
CC Attribution-Noncommercial-Share Alike 3.0 Unported
Driven by DokuWiki