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A dish of mass $10\ g$ is kept floating horizontally in the air by firing bullets, each of mass $5\ g$, with the same velocity, at the rate of $10$ bullets per second and the bullets rebound with the same speed in opposite direction. The velocity of each bullet at the time of impact is ___(Take $g=10\ m\ s^{-2}$)

My thoughts:-

dish apply $F=0.1\ N$ downwards, so we need to provide bullets velocity in the upward direction in an amount so that it can apply an equal force $(F=0.1\ N)$.

so the mass of bullets times its velocity $(v)$ $=5 \times 10 \times v \ gram =0.05v \ kg$ has to be equall to $F=0.1 \ N$ $\implies v=2 m/s^{-1}$. But this didn't match any of options.

Also if the force applied by dish on bullets and vice versa is same, so they should cancel each other, then how bullets can come back with same speed, it should transfer all its momentum to dish to hold it there.

Please point out where my concepts are lacking. And please clear the last assertion of mine.

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mass times velocity is not a force. What is matter is the change of momentum of a bullet before and after the collision. The rate of change $Δp/Δt$ of momentum of all bullets that collide with dish at an interval $Δt$, gives the mean force that bullets exerts at dish.

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  • $\begingroup$ then is $1 \ m/s^{-1}$ is correct? $\endgroup$ – mnulb Dec 3 '16 at 9:15
  • $\begingroup$ v=g dish_mass/(2 bullet rate x bullet_mass $\endgroup$ – user14178 Dec 3 '16 at 10:16
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Also if the force applied by dish on bullets and vice versa is same, so they should cancel each other, then how bullets can come back with same speed, it should transfer all its momentum to dish to hold it there.

This action-reaction pair of forces act on different bodies. One acts on the dish, the other on the bullets. They will only cancel if they act on the same body.

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