A 0.70 kg ball moving horizontally at 5.0 m/s strikes a vertical wall and rebounds with speed 2.0 m/s. What is the magnitude of the change in its linear momentum

4.9 kg. m/s.

Explanation:

Given that

mass, m = 0.7 kg

Initial speed, u = 5 m/s (Towards + x direction)

Final speed, v = - 2 m/s (Towards - x direction)

We know that linear momentum is given as

P = Mass x velocity

Change in the linear momentum ΔP will be

ΔP = m (v - u)

Now by putting the values in the above equation, we get'

ΔP = 0.7 (-2 - 5) kg. m/s

ΔP = - 4.9 kg. m/s

The magnitude of the change in the linear momentum will be 4.9 kg. m/s.