A 64.0-kg person holding two 0.900-kg bricks stands on a 2.00-kg skateboard. Initially, the skateboard and the person are at rest. The person now throws the two bricks at the same time so that their speed relative to the person is 19.0 m/s. What is the recoil speed of the person and the skateboard relative to the ground, assuming the skateboard moves without friction?

0.518 m/s

Explanation:

To solve this question, we would use the Law of conservation of momentum.

It is stated that the skateboard and the person were at rest initially, this means that their initial momentum is 0. Since they aren't moving.

And from the law if conservation of momentum, we know that initial momentum must be equal to final momentum.

The final momentum of the bricks will be 2* 0.9 * 19 = 34.2 kg m/s.

This means that the momentum of the person and that of the skateboard has to be 34.2 kg m/s also, although, it will be in the opposite direction.

Mass of the person plus that of the skateboard is

64 + 2 = 66 kg,

If we divide the momentum of the person and that of the skateboard by the mass of the both of them, we get the speed at which they are moving after the bricks are thrown

Recoil speed = (34.2 kg m/s)/(66 kg) = 0.518 m/s.

the speed of the cart after 3 seconds of low fan speed is ⇒ 54.0 cm/s.

the speed of the cart after 5 seconds of medium fan speed is ⇒ 120.0 cm/s.

the speed of the cart after 2 seconds of high fan speed is ⇒ 64.0 cm/s. 

just did the lab.