Starting with an initial speed of 4.54 m/s at a height of 0.337 m, a 2.03-kg ball swings downward and strikes a 4.97-kg ball that is at rest, as the drawing shows.
(a) using the principle of conservation of mechanical energy, find the speed of the 2.03-kg ball just before impact.
(b) assuming that the collision is elastic, find the velocity (magnitude and direction) of the 2.03-kg ball just after the collision.
(c) assuming that the collision is elastic, find the velocity (magnitude and direction) of the 4.97-kg ball just after the collision.
(d) how high does the 2.03-kg ball swing after the collision, ignoring air resistance

a) 5.2170 m/s

b) 2.1911 m/s, to the left

c) 3.0259 m/s, to the right

d) 0.2449 m

Explanation:

Given parameters:

a) Here, the conservation of energy principle can be used.

where

So,

b) Here, the conservation of momentum principle can be used.

Here, our new will be   found in part (a).

(-) sign means that the 2.03 kg ball after collision will bounce back in the left direction.

c) Here, we can use the conservation of momentum principle again.

. Here, will be   found in part (a) again.

(+) sign means that the 4.97 kg ball after collision will swing to the right direction.

d) Here, we can use the conservation of energy principle again.

where and

So,

d is the answer i came up with