A block of mass 1 kg moving with a speed of $4 m s^{- 1}$ , collides with another block of mass 2kg which is at rest. The lighter block comes to rest after collision. The speed of the heavy block in the same direction is:

2 m s^{- 1}

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

0.67 m s^{- 1}

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

0.5 m s^{- 1}

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

0.3 m s^{- 1}

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

Open in App

Solution

The correct option is A $2 m s^{- 1}$
$m_{1} = 1 k g u_{1} = 4 m / s m_{2} = 2 k g u_{2} = ?$
using principal of linear momentum conservation:-
$m_{1} u_{1} {= m}_{2} u_{2}$
$1 \times 4 = 2 \times u_{2}$
$u_{2} = 2 m / s$

Suggest Corrections

Similar questions

Q. A block of mass 1 kg moving with a speed of

4 m s^{- 1}

, collides with another block of mass 2kg which is at rest. The lighter block comes to rest after collision. The loss in the K.E of the system

A block of mass m moving with speed v collides with another block of mass 2 m at rest. The lighter block comes to rest after collision. The value of coefficient of restitution is

Q. A block of mass 1 kg moving with a speed of

2 {m s}^{- 1}

collides with and sticks to another block of mass 2 kg which was initially at rest. After the collision

Q. A block of mass

m

moving at a velocity

v

collides with another block of mass

2 m

at rest. The lighter block comes to rest after collision. Find the coefficient of restitution.

Q. A block of mass

m

moving at a speed

v

collides with another block of mass

2 m

at rest. The lighter block comes to rest after collision. What is the coefficient of restitution.

Join BYJU'S Learning Program

A block of mass 1 kg moving with a speed of4ms−1, collides with another block of mass 2kg which is at rest. The lighter block comes to rest after collision. The speed of the heavy block in the same direction is:

The correct option is A 2ms−1m1=1kgu1=4m/sm2=2kgu2=?using principal of linear momentum conservation:-m1u1=m2u21×4=2×u2u2=2m/s

A block of mass 1 kg moving with a speed of $4 m s^{- 1}$ , collides with another block of mass 2kg which is at rest. The lighter block comes to rest after collision. The speed of the heavy block in the same direction is:

The correct option is A $2 m s^{- 1}$
$m_{1} = 1 k g u_{1} = 4 m / s m_{2} = 2 k g u_{2} = ?$
using principal of linear momentum conservation:-
$m_{1} u_{1} {= m}_{2} u_{2}$
$1 \times 4 = 2 \times u_{2}$
$u_{2} = 2 m / s$