Question

A 72.0 kg stuntman jumps from a moving car to a 2.50 kg skateboard at rest. If the velocity of the car is 15.0 m/s to the east when the stuntman jumps, what is the final velocity of the stuntman and the skateboard?
A 0.521 m/s to the east
B 14.5 m/s to the east
C 15.5 m/s to the east
D 432 m/s to the east

Answer 1

Answer:

B 14.5 m/s to the east

Explanation:

We can solve this problem by using the law of conservation of momentum.

In fact, if the system is isolated, the total momentum of the system must be conserved.

Here the total momentum before the stuntman reaches the skateboard is:

$p_i = Mv$

where

M = 72.0 kg is the mass of the stuntman

v = 15.0 m/s is his initial velocity (to the east)

The total momentum after the stuntmen reaches the skateboard is:

$p_f = (M+m)v'$

where

m = 2.50 kg is the mass of the skateboard

v' is the final velocity of the stuntman and the skateboard

Since momentum must be conserved, we have

$p_i = p_f\\Mv=(M+m)v'$

And solvign for v',

$v'=\frac{Mv}{M+m}=\frac{(72.0)(15.0)}{(72.0+2.50)}=14.5 m/s$

And since the sign is the same as v, the direction is the same (to the east).