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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 Ns



Given,

F=6002×105t

The time when this force becomes zero

6002×105t=0

Or, t=3×103s

Time taken by the bullet to leave the barrel is 3 milliseconds.

We know,

Impulse = F.idt=(6002×105t)dt

Or, J=[600t(2×105t22)]

Now putting values of t in above equation

J=600×(3×103)2×105×(3×103)22

After solving

J=0.9Ns

Hence, the impulse is 0.9 Ns.




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