A 1.40 kg skateboard is coasting along the pavement at a speed of 6.99 m/s when a 0.660 kg cat drops..........

A 1.40 kg skateboard is coasting along the pavement at a speed of 6.99 m/s when a 0.660 kg cat drops from a tree vertically downward onto the skateboard.

What is the speed of the skateboard-cat combination? Answer in units of m/s

According to the law of conservation of linear momentum, the total linear momentum of a closed system before a collision is equal to that after a collision. Since the cat falls directly along the normal to the skateboard, the horizontal component of its initial speed is 0.

Linear momentum before the collision, ρ1
= (m1u1)+(m2u2), where m1 is the mass of the skateboard, u1 is the initial speed of the skateboard, m2 is the mass of the cat and u2 is the horizontal component of the cat's speed.

ρ1= (1.40)(6.99)+0
= 9.786kgm/s

Linear momentum after the collision, ρ2
= (m1+m2)v, where v is the final speed of the cat and the skateboard

(m1+m2)v= 9.786
2.06v= 9.786
v= 9.786/2.06
= 4.75m/s

I hope this helps and feel free to send me an e-mail if you have any doubts!