A 3-kg skateboard is rolling down the sidewalk at 4 m/s when it collides with a 1-kg skateboard that was initially at rest. If the 1-kg skateboard moves off at 6 m/s after the collision, what is the new velocity for the 3-kg skateboard?

2 m/s

Explanation:

Parameters given:

Mass of first skateboard, m = 3 kg

Initial speed of first skateboard, u = 4 m/s

Mass of second skateboard, M = 1 kg

Initial speed of second skateboard, U = 0 m/s

Final speed of second skateboard, V = 6 m/s

Using the principle of the conservaton of momentum, the total initial momentum is equal to the total final momentum.

Momentum is the product of mass and velocity. This implies that:

m*u + M*U = m*v + M*V

(3*4) + (1*0) = (3*v) + (1*6)

12 + 0 = 3v + 6

=> 3v = 12 - 6

3v = 6

v = 6/3 = 2 m/s

The final speed of the 3 kg skateboard is 2 m/s

2m/s

Explanation:

Using the law of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the sum of their momentum after collision.

Momentum = mass * velocity.

Before collision:

Momentum of 3kg body moving at 4m/s = 3*4

= 12kgm/s

Momentum of 1kg skateboard initially at rest = 1*0

= 0kgm/s (the velocity 0m/s because the body is initially at rest)

After collision:

1-kg skateboard moves off at 6m/s. Its momentum after collision will be;

1*6 = 6kgm/s

For the 3kg body moving with an unknown velocity v, its final momentum will be;

3*v = 3v

Applying the law as stated above;

12+0 = 6+3v

12 = 6+3v

12-6 = 3v

6 = 3v

v = 6/3

v = 2m/s

This means that the new velocity of the 3kg skateboard after collision is 2m/s