A person A moving in a car with a speed of $10 m/s$ drops a stone of mass $10 gm$ . A person B on the ground measures the momentum of the stone 1 sec after the stone is dropped. Find the momentum ( $kg m/s$ ) of the stone as measured by B. (Take $g = 10 {m/s}^{2}$ )

0.707

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0.14

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1.4

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7.07

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Solution

The correct option is B $0.14$
As stone is initially in the car, its velocity in the horizontal direction $v_{x} = u_{x} = 10^i m/s$
In the vertical direction: $v_{y} = u_{y} + g t = 10^j m/s$ .
Hence net momentum $m (v_{x}^i + v_{y}^j) = 10 \times 10^{- 3} \times 10 \sqrt{2} = 0.14 kg m/s$

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A person A moving in a car with a speed of 10 m/s drops a stone of mass 10 gm. A person B on the ground measures the momentum of the stone 1 sec after the stone is dropped. Find the momentum (kg m/s) of the stone as measured by B. (Take g=10 m/s2)

The correct option is B 0.14As stone is initially in the car, its velocity in the horizontal direction vx=ux=10^i m/s In the vertical direction: vy=uy+gt=10^j m/s . Hence net momentum m(vx^i+vy^j)=10×10−3×10√2=0.14 kg m/s

A person A moving in a car with a speed of $10 m/s$ drops a stone of mass $10 gm$ . A person B on the ground measures the momentum of the stone 1 sec after the stone is dropped. Find the momentum ( $kg m/s$ ) of the stone as measured by B. (Take $g = 10 {m/s}^{2}$ )

The correct option is B $0.14$
As stone is initially in the car, its velocity in the horizontal direction $v_{x} = u_{x} = 10^i m/s$
In the vertical direction: $v_{y} = u_{y} + g t = 10^j m/s$ .
Hence net momentum $m (v_{x}^i + v_{y}^j) = 10 \times 10^{- 3} \times 10 \sqrt{2} = 0.14 kg m/s$