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?

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

no

sodium is solid at rt

we need to know the time interval during which the force acts. impulse of force = momentum change