You visited us 1 times! Enjoying our articles? Unlock Full Access!

Partially Inelastic Collision

A particle of...

Question

A particle of mass $m_{1}$ moving with velocity v in a positive direction collides elastically with a mass $m_{2}$ moving in opposite direction also at velocity v. If $m_{2}$ >> $m_{1}$ , then :

the velocity of

m_{1}

immediately after collision is nearly 3v

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

the change in momentum of

m_{1}

is nearly

4 m_{1} v

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

the change in kinetic energy of

m_{1}

is nearly

4 m v_{2}

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

all of the above

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

Open in App

Solution

The correct option is D all of the above
Let the velocity of mass $m_{1}$ after the collision be $v_{1}$ in positive direction.
Given : $u_{1} = v$ $u_{u} = - v$ $e = 1$ $m_{2} >> m_{1}$

Using $v_{1} = \frac{(m_{1} - e m_{2}) u_{1} + (1 + e) m_{2} u_{2}}{m_{1} + m_{2}}$

$v_{1} = \frac{(m_{1} - (1) m_{2}) v + (1 + 1) m_{2} (- v)}{m_{1} + m_{2}} = \frac{m_{1} v - 3 m_{2} v}{m_{1} + m_{2}}$
For $m_{2} >> m_{1}$ $⟹ v_{1} = - 3 v$ (-ve sign represents $m_{1}$ recoils)
Now change in momentum $Δ P = m_{1} v_{2} - m_{1} v$
$Δ P = m_{1} (- 3 v) - m_{1} v = - 4 m_{1} v$ $⟹ | Δ P | = 4 m_{1} v$

Change in kinetic energy of $m_{1}$ $Δ E = \frac{1}{2} m_{1} v_{1}^{2} - \frac{1}{2} m_{1} v^{2}$
$Δ E = \frac{1}{2} m_{1} (3 v)^{2} - \frac{1}{2} m_{1} v^{2} = 4 m_{1} v^{2}$

Suggest Corrections

Similar questions

Q. A mass of

m_{1}

is a moving with velocity

v

. It collides head on elastically with a stationary ball of mass

m_{2}

. The velocity of ball becomes

\frac{v}{3}

after collision,then the value of the ratio

\frac{m_{2}}{m_{1}}

is:

Q. Object

A

of mass

m

, is moving at a velocity

v_{1}

to the right. It collides and sticks to object

B

of mass

m_{2}

moving in the same direction as object

A

with velocity

v_{2}

. After collision, both the objects have the same velocity

\frac{1}{2} (v_{1} + v_{2})

. What is the relationship between

m_{1}

and

m_{2}

? [

v_{1}

and

v_{2}

are not equal]

Q. A particle of mass

m_{1}

and moving with velocity

v_{1}

collides head on with another stationary particle of mass

m_{2}

elastically. If after the collision their velocities are

v_{1}

and

v_{2}

then under the condition

m_{1} = m_{2}

their values will be :

Q. A particle of mass

m_{1}

moving with velocity v collides with a mass

m_{2}

at rest, then they get embedded. Just after collision, velocity of the system :

Q. A particle of mass

m_{1}

moving with a velocity of

5 m / s

collides head on with a stationary particle of mass

m_{2}

. After collision both the particle move with a common velocity of

4 m / s

, then the value

m_{1} / m_{2}

is:

Join BYJU'S Learning Program

A particle of mass m1 moving with velocity v in a positive direction collides elastically with a mass m2 moving in opposite direction also at velocity v. If m2 >> m1, then :

A particle of mass $m_{1}$ moving with velocity v in a positive direction collides elastically with a mass $m_{2}$ moving in opposite direction also at velocity v. If $m_{2}$ >> $m_{1}$ , then :