You visited us 190 times! Enjoying our articles? Unlock Full Access!

Finding Average Velocity

A 2.5 kg ba...

Question

A $2.5 k g$ ball is dropped onto a concrete floor. It strikes the floor with a momentum of $20 k g m / s$ and bounces away from the floor with a momentum of $16 k g m / s$ .What is the magnitude of change of momentum of the ball?

4 k g . m / s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

8 k g m / s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

32 k g m / s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

36 k g m / s

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

40 k g m / s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $36 k g m / s$
Let the upward direction to be positive.
Initial momentum $P_{i} = - 20 k g m / s$
Final momentum $P_{f} = 16 k g m / s$
Change in momentum $Δ P = P_{f} - P_{i} = 16 - (- 20) = 36 k g m / s$

Suggest Corrections

Similar questions

5 kg m/s

p

\frac{1}{2}

1

25

10

2 m / s^{2}

(g = 10 m / s^{2})

100 g

2 m

1.5 m

(a)

(b)

10^{- 8} s

Join BYJU'S Learning Program