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Question

# A bullet is fired from a gun. The force on the bullet is given by F = 600 - 2 × 105 t, where F is in Newton and t in seconds. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet

A
9 Ns
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B
Zero
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C
0.9 Ns
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D
1.8 Ns
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Solution

## The correct option is (C) 0.9 NsGiven, F=600−2×105t The time when this force becomes zero600−2×105t=0 Or, t=3×10−3s Time taken by the bullet to leave the barrel is 3 milliseconds. We know, Impulse = F.idt=(600−2×105t)dt Or, J=[600t−(2×105t22)] Now putting values of t in above equation J=⎡⎣600×(3×10−3)−⎛⎝2×105×(3×10−3)22⎞⎠⎤⎦ After solving J=0.9Ns Hence, the impulse is 0.9 Ns.

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