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On a frictionless track, a trolley moves with a speed of $36 \mathrm{~km} / \mathrm{h}$ with a mass of $\mathbf{2 0 0} \mathrm{Kg}$. A child whose mass is 20 kg runs on the trolley with a speed of $4 \mathbf{m} \mathbf{s}^{1}$ from one end to other which is $20 \mathrm{~m}$. The speed is relative to the trolley in the direction opposite to its motion. Find the final speed of the trolley and the distance the trolley moved from the time the child began to run.
On a frictionless track, a trolley moves with a speed of $36 \mathrm{~km} / \mathrm{h}$ with a mass of $\mathbf{2 0 0} \mathrm{Kg}$. A child whose mass is 20 kg runs on the trolley with a speed of $4 \mathbf{m} \mathbf{s}^{1}$ from one end to other which is $20 \mathrm{~m}$. The speed is relative to the trolley in the direction opposite to its motion. Find the final speed of the trolley and the distance the trolley moved from the time the child began to run.
CBSE | Class 11 | NCERT | Physics | Work, Energy, and Power
On a frictionless track, a trolley moves with a speed of $36 \mathrm{~km} / \mathrm{h}$ with a mass of $\mathbf{2 0 0} \mathrm{Kg}$. A child whose mass is 20 kg runs on the trolley with a speed of $4 \mathbf{m} \mathbf{s}^{1}$ from one end to other which is $20 \mathrm{~m}$. The speed is relative to the trolley in the direction opposite to its motion. Find the final speed of the trolley and the distance the trolley moved from the time the child began to run.

Mass is given as $m=200 \mathrm{Kg}$

Speed is given as $v=36 \mathrm{~km} / \mathrm{h}=10 \mathrm{~m} / \mathrm{s}$

Mass of boy is given as $=20 \mathrm{Kg}$

Initial momentum will be,

$(M+m) v$

$=(200+20) \times 10$

$=2200 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$

$v^{\prime}$ is final velocity of the trolley

Final velocity of boy can be calculated as,

$=M v^{\prime}+m\left(v^{\prime}-4\right)$

$=200 v^{\prime}+20 v^{\prime}-80$

$=220 v^{\prime}-80$

According to law of conservation of energy, we have,

Initial momentum $=$ final momentum

$2200=220 v^{\prime}-80$

$v^{\prime}=\frac{2280}{220}=10.36 \mathrm{~m} / \mathrm{s}$

Length $\mathrm{I}=20 \mathrm{~m} / \mathrm{s}$

$v^{\prime \prime}=4 \mathrm{~m} / \mathrm{s}$

$t=\frac{20}{4}=5 \mathrm{~s}$

Distance moved by the trolley $=v^{\prime \prime} \mathrm{xt}=10.36 \times 5=51.8 \mathrm{~m}$

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