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 14.5m/s to the east
C 15.5 m/s to the east
D 432 m/s to the eas

Answer 1

The final velocity of the skateboard is 14.5 m/s

Step-by-step explanation:

Given:

Mass of the man, m₁ = 72 kg

Velocity of the car, v₁ = 15 m/s

Mass of the skateboard, m₂ = 2.5 kg

Velocity of the skateboard = 0 (at rest)

Final velocity, V = ?

When the stuntman jumps onto the skateboard, the momentum remains conserved.

Thus,

m₁v₁ + m₂ X v₂ = (m₁ + m₂) V

On substituting the value we get:

`72 X 15 + m_2 X 0 = (72 + 2.5 ) V\\\\1080 + 0 = (74.5) V\\\\\\`

`V = (1080)/(74.5) \\\\V = 14.49\\\\V = 14.5 m/s`

Thus, the final velocity of the skateboard is 14.5 m/s