JEE MAIN - Physics (2022 - 28th June Morning Shift - No. 23)
A man of 60 kg is running on the road and suddenly jumps into a stationary trolly car of mass 120 kg. Then, the trolly car starts moving with velocity 2 ms$$-$$1. The velocity of the running man was ___________ ms$$-$$1, when he jumps into the car.
Answer
6
Explanation
Total momentum of (man + trolley) system is always conserved
Initially man was moving with velocity v1 and trolley was at rest, finally both were moving with velocity 2 ms$$-$$1 after man jumps on the trolley.
So,
$ \Rightarrow \quad m_{1} v_{1}+0=\left(m_{1}+m_{2}\right) v_{2} $
$ \text { Here, } m_{1}=\text { mass of man }=60 \mathrm{~kg} $
$ m_{2}=\text { mass of trolley }=120 \mathrm{~kg} $
$ v_{1}=\text { speed of } \text { man } $
$ v_{2}=\text { speed of man and trolley }=2 \mathrm{~m} / \mathrm{s} $
$ \Rightarrow 60 \times v_{1}=(60+120) \times 2 $
$ \Rightarrow v_{1}=\frac{(60+120) \times 2}{60}=6 \mathrm{~m} / \mathrm{s} $
Initially man was moving with velocity v1 and trolley was at rest, finally both were moving with velocity 2 ms$$-$$1 after man jumps on the trolley.
So,
$ \Rightarrow \quad m_{1} v_{1}+0=\left(m_{1}+m_{2}\right) v_{2} $
$ \text { Here, } m_{1}=\text { mass of man }=60 \mathrm{~kg} $
$ m_{2}=\text { mass of trolley }=120 \mathrm{~kg} $
$ v_{1}=\text { speed of } \text { man } $
$ v_{2}=\text { speed of man and trolley }=2 \mathrm{~m} / \mathrm{s} $
$ \Rightarrow 60 \times v_{1}=(60+120) \times 2 $
$ \Rightarrow v_{1}=\frac{(60+120) \times 2}{60}=6 \mathrm{~m} / \mathrm{s} $
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