6.28 A trolley of mass 200 kg moves with a uniform speed of 36 km/h on a frictionless track. A child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of 4 m s–1 relative to the trolley in a direction opposite to the its motion, and jumps out of the trolley. What is the final speed of the trolley? How much has the trolley moved from the time the child begins to run?

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6.28 A trolley of mass 200 kg moves with a uniform speed of 36 km/h on a frictionless track. A child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of 4 m s^–1 relative to the trolley in a direction opposite to the its motion, and jumps out of the trolley. What is the final speed of the trolley? How much has the trolley moved from the time the child begins to run?

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Payal Gupta | Contributor-Level 10

4 months ago

6.28 Mass, m = 200 kg

Speed, v = 36 km/h = 10 m/s

Mass of the boy, M = 20 kg

Initial momentum = (M + m)v = (20 + 200) x 10 kg-m/s = 2200 kg-m/s

If v' is the final velocity of the trolley, then

The final momentum = (M+m) x v' – M x 4= 220v'-80

According to the law of conservation of energy,

Initial momentum = final momentum

2200 = 220v'-80

V' = 10.36 m/s

Time required by the boy to travel 10m = 10/4 = 2.5 s

Distance travel by trolley in 2.5 s = 10.36 x 2.5 m = 25.9 m

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(b) conserving energy between “O” ans ”A”

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$\frac{{(v')}^{2}}{2} = \frac{v^{2}}{2} = - g h$

(v’)²=v²-2gh = v^’= $\sqrt[]{v^{2} - 2 g h}$ ……….1

Let speed after emerging be v₁then

=1/2mv₁²=1/2[1/2mv’²]

1/2m(v₁)²=1/4m(v’)²=1/4m[v²-2gh]

V₁= $\sqrt[]{\frac{v^{2}}{2} - g h}$ ………….2

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$\frac{v'}{v_{1}} = \frac{\sqrt[]{v^{2} - 2 g h}}{\sqrt[]{v^{2} - 2 g h / \sqrt[]{2}}} = \sqrt[]{2}$

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