You are a world-famous physicist-lawyer defending a client who has been charged

Question

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: 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. with only a slight hesitation, you turn confidently to the jury and proclaim, "my client's gun did not fire these shots!" how are you going to convince the jury and the judge?

Explanation / Answer

m v = (m + M) V   conservation of momentum
m = mass of bullet and M = mass of chair
V = m v / (m + M)     speed of chair after collision
You can use M = M + m since m is small compared to M
V = m v / M
1/2 M V^2 = u M g x    where x is distance chair slides and u = friction
u = V^2 / (2 * g * x)   solving for u
u = (m v / M)^2 / (2 * g * x)    substituting for V
u = (.01 * 450 / 20)^2 / ( 2 * 9.8 * .03) = .086
Since the coefficient of friction of dry wood on dry wood ranges from .25 to .50
your client's gun could not have fired the bullet.

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