You visited us 3 times! Enjoying our articles? Unlock Full Access!

Partially Inelastic Collision

A body of mas...

Question

A body of mass $m_{1}$ collides elastically with another body of mass $m_{2}$ at rest. If the velocity of $m_{1}$ after collision becomes $\frac{2}{3}$ times its initial velocity, the ratio of their masses is

$1 : 5$

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

$5 : 1$

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

$5 : 2$

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

$2 : 5$

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

Open in App

Solution

The correct option is B
$5 : 1$

Step 1. Given data:
Mass of first body, $m_{1}$ with velocity $v_{1}$
Mass of second body, $m_{2}$ with velocity, $v_{2} = 0$
After the collision, the final velocity of the first body, $v_{1}' = \frac{2}{3} v_{1}$
Step 2. Applying the conservation of the momentum
Momentum is the product of mass and velocity.
According to momentum conservation, the initial momentum of the system before the collision is equal to the final momentum of the system after the collision.
Using momentum conservation,
$m_{1} v_{1} + m_{2} v_{2} = m_{1} v_{1}' + m_{2} v_{2}'$ …… $(1)$ [ $v_{2}^{'}$ is the final velocity of the second body]
$m_{1} v_{1} + 0 = m_{1} v_{1}^{'} + m_{2} v_{2}^{'}$ [ from $(1)$ ] [ $v_{2} = 0$ ]
$m_{1} v_{1} = m_{1} \times \frac{2}{3} v_{1} + m_{2} v_{2}^{'}$ [ $v_{1}^{'} = \frac{2}{3} v_{1}$ ]
$m_{2} v_{2}^{'} = m_{1} v_{1} - \frac{2}{3} m_{1} v_{1}$
$m_{2} v_{2}' = \frac{1}{3} m_{1} v_{1}$ …… $(3)$
Step 3. Applying conservation of kinetic energy
Kinetic energy is equal to half of the product of the mass and the square of the velocity.
According to energy conservation, the energy before the collision is equal to the energy after the collision.
Kinetic energy conservation,
$\frac{1}{2} m_{1} {v_{1}}^{2} + \frac{1}{2} m_{2} {v_{2}}^{2} = \frac{1}{2} m_{1} v_{1}'^{2} + \frac{1}{2} m_{2} v_{2}'^{2}$ ……. $(2)$
$\frac{1}{2} m_{1} v_{1}^{2} + 0 = {\frac{1}{2} m_{1} (\frac{2}{3} v_{1})}^{2} + \frac{1}{2} m_{2} v_{2}^{' 2}$ [ from $(2)$ ]
$\frac{1}{2} m_{2} v_{2}^{' 2} = \frac{1}{2} m_{1} v_{1}^{2} - \frac{4}{18} m_{1} v_{1}^{2}$
$m_{2} v_{2}'^{2} = \frac{5}{9} m_{1} v_{1}^{2}$ ……. $(4)$
Dividing equation $(4)$ by equation $(3)$ , we get
$v_{2}' = \frac{5}{3} v_{1}$
Now, putting the value of $v_{2}'$ in equation $(3)$ ,
$\frac{1}{3} m_{1} v_{1} = m_{2} \frac{5 v_{1}}{3}$
$\frac{m_{1}}{m_{2}} = \frac{5}{1}$
Hence, option (B) is correct.

Suggest Corrections

Similar questions

Q. A body of mass

m_{1}

moving with a velocity

3 m / s

collides with another body at rest of mass

m_{2}

. After collision the velocities of the two bodies are

2 m / s

and

5 m / s

, respectively, along the direction of motion of

m_{1}

. The ratio

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

Q. A body of mass

m_{1}

moving with a velocity 10 m/s collides elastically with another body at rest of mass

m_{2}

. After collision the velocites of the two bodies are 2 m/s and 5 m/s respectively along the direction of motion of

m_{1}

. The ration of

\frac{m_{1}}{m_{2}} i s

Q. A body of mass

m_{1}

collides head on elastically with a stationary body of mass

m_{2}

. If velocities of

m_{1}

before and after the collision are

v

and

- v / 3

respectively then the value of

m_{1} / m_{2}

Q. An object of mass

m_{1}

moving with a velocity V undergoes a head on collision with another object of mass

m_{2}

at rest. After collision body with mass

m_{1}

comes to the state of rest. Determine the velocity of the body with mass

m_{2}

after collision.

Q. A body of mass

m_{1}

is moving with a velocity

3 m s^{- 1}

collides with another body at rest of mass

m_{2}

. After collision the velocities of the two bodies are

2 m s^{- 1}

and

5 m s^{- 1}

respectively along the direction of motion of

m_{1}

. The ratio

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

Join BYJU'S Learning Program

A body of mass m1 collides elastically with another body of mass m2 at rest. If the velocity of m1 after collision becomes23 times its initial velocity, the ratio of their masses is

A body of mass $m_{1}$ collides elastically with another body of mass $m_{2}$ at rest. If the velocity of $m_{1}$ after collision becomes $\frac{2}{3}$ times its initial velocity, the ratio of their masses is