The final velocity of the skateboard is 14.5 m/s
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:
[tex]72 X 15 + m_2 X 0 = (72 + 2.5 ) V\\\\1080 + 0 = (74.5) V\\\\\\[/tex]
[tex]V = \frac{1080}{74.5} \\\\V = 14.49\\\\V = 14.5 m/s[/tex]
Thus, the final velocity of the skateboard is 14.5 m/s
