(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

You are a world-famous physicist-lawyer defending a client who has been charged with murder. It is alleged that your client, Mr. Smith, shot the victim, Mr. Wesson. The detective who investigated the scene of the crime found a second bullet, from a shot that missed Mr. Wesson, that had embedded itself into a chair. You arise to cross-examine the detective. You: In what type of chair did you find the bullet? Det: A wooden chair. You: How massive was this chair? Det: It had a mass of 20.0 kg. You: How did the chair respond to being struck with a bullet? Det: It slid across the floor. You: How far? Det: Three centimeters. The slide marks on the dusty floor are quite distinct. You: What kind of floor was it? Det: A wood floor, very nice oak planks. You: What was the mass of the bullet you retrieved from the chair? Det: Its mass was 10 g. You: And how far had it penetrated into the chair? Det: A distance of 4.00 cm. You: Have you tested the gun you found in Mr. Smith's possession? Det: I have. You: What is the muzzle velocity of bullets fired from that gun? Det: The muzzle velocity is 450 m/s. You: And the barrel length? Det: The gun has a barrel length of 62 cm.

2. Relevant equations

3. The attempt at a solution

I found the force on chair by the bullet if the bullet is going 450 m/s

0=(450)^2 +2a(.04m)

a= -2531250 m/s^2

F= (.01kg)(-2531250m/s^2)

So force by bullet on chair is 25312.5 N

This force acts this time:

for 0=(450m/s) + (-2531250m/s^2)(t)

t= 1.78 * 10^-4 s

Force of friction on chair : (20.01kg)(9.80)(.20) = 39.2 N

sum forces on x = Fb-kinetic friction = 25312.5N-39.2N= 25273.3 N

acceleration of chair= (20.01kg)(a)= 25273.3N a=1263 m/s^2

Impulse-momentum theorem: Impulse plus initial momentum equals final momentum

(25273.3N)(1.78*10^-4s) + 0 = (20.01kg)v

v=.225m/s this is the velocity of the chair when the bullet stops moving

distance it has gone during the impulse:

(.225m/s)^2= (0m/s)^2 + 2(1263 m/s^2)x

x=2.00*10^-5 m

after this: Force of friction = 39.2 N

(20.01kg)(a)=(-39.2N) a=-1.95 m/s^2

distance traveled during this time

(0m/s)^2=(.225 m/s)^2 + 2(-1.95m/s^2)x

x=.013 m =1.29 cm.

Is this correct, or am I making this way too complicated? If it is not correct, could you please explain how you would have done it?

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# Homework Help: World Famous Physicist Lawyer

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