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

Law of Conservation of Energy

A mass m1 mov...

Question

A mass $m_{1}$ moving with initial velocity $u$ strikes head on to another mass $m_{2}$ kept at rest as shown in figure. After collision if $| v_{1} | < | u |$ , then

m_{1} > m_{2}

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

m_{1} < m_{2}

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

m_{1} = m_{2}

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

There is no relation between

m_{1}

and

m_{2}

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

Open in App

Solution

The correct option is B $m_{1} < m_{2}$

Using momentum conservation,
$p_{i n i t i a l} = P_{f i n a l}$
$m_{1} u = m_{2} v_{2} - m_{1} v_{1} . . . (i)$

For elastic collision,
$velocity of approach = velocity of separation$
$u = v_{2} - (- v_{1}) = v_{2} + v_{1} \Rightarrow v_{2} = u - v_{1} . . . . (i i)$

from $(i)$ and $(i i)$
$m_{1} u = m_{2} (u - v_{1}) - m_{1} v_{1} v_{1} = \frac{m_{2} - m_{1}}{m_{1} + m_{2}} u$

since, we already assumed $v_{1}$ is in opposite direction of $u$
$∴ v_{1}$ must be greater than zero
$v_{1} = \frac{m_{2} - m_{1}}{m_{1} + m_{2}} u > 0 \Rightarrow m_{2} - m_{1} > 0 \Rightarrow m_{2} > m_{1}$

Suggest Corrections

Similar questions

m_{1}

u

m_{2}

| v_{1} | < | u |

m_{1}

m_{2}

m_{1}

m_{2}

m_{1}

m_{2}

m_{1}

m_{2}

m_{1}

v_{1}

m_{2}

v_{1}

v_{2}

m_{1} = m_{2}

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program