Question

A block of mass 22 kg is sliding along the ice at constant speed 5.0 m/s. Just ahead of it is a block of mass 29 kg sliding in the

same direction at constant speed 4.6 m/s. When the two blocks collide, the 29-kg block travels at a new speed of 7.2 m/s. What is
the new speed of the 22-kg block?

Answer 1

The new speed of 22 kg block is 1.57 m/s

Step-by-step explanation:

Given-

Mass,
`m_(1)`= 22 kg

speed,
`v_(1)` = 5 m/s

`m_(2)` = 29 kg

`v_{2i` = 4.6 m/s

New speed of
`m_(2)`,
`v_(2f)` = 7.2 m/s

New speed of
`m_(1)`,
`v_(1f)` = ?

This is the case of elastic collision.

So,

`M_(1)` X
`V_(1i)`+
`M_(2)` X
`V_(2i)` =
`M_(1)` X
`V_(1f)` +
`M_(2)` X
`V_(2f)`

22 X 5 + 29 X 4.6 = 22 X
`V_(1f)` + 29 X 7.2

22
`V_(1f)` + 208.8 = 243.4

22
`V_(1f)`= 34.6

`V_(1f)`= 1.57 m/s

Therefore, the new speed of 22 kg block is 1.57 m/s