A particle of mass ' $m$ ' moving with velocity ' $v$ ' collides with stationary particle of mass $2 m$ . Then the speed of the system after collision is :

2 v

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

\frac{v}{2}

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

3 v

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

\frac{v}{3}

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

Open in App

Solution

The correct option is D $\frac{v}{3}$
Mass of first particle is $m_{1} = m$ and its velocity before collision $u_{1} = v$

Mass of second particle is $m_{2} = 2 m$ and its velocity before collision $u_{2} = 0$

Let the velocity of combined system just after collision be $V$ .

Using conservation of linear momentum :

$m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) V$

$∴ (m) (v) + (2 m) (0) = (m + 2 m) V$

$⟹$ $V = \frac{v}{3}$

Suggest Corrections

Similar questions

45^{o}

\frac{v}{2}

A particle of mass m moving with velocity v strikes a stationary particle of mass 2m and sticks to it. The speed of the system will be

m

v

v

2 m

m_{1}

m_{2}

Join BYJU'S Learning Program