A bullet is fired from a gun. The force on the bullet is given by F=600−2×105t, where F is in newtons 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
9Ns
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B
Zero
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C
0.9Ns
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D
1.8Ns
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Solution
The correct option is C0.9Ns Given F=600−2×105t ′F′ is zero ⇒0=600−2×105t /63/20001000=t t=31000 Now impulse =t∫0Fdt ⇒t∫0(600−2×105t)dt =600t−105t2 =600×31000−105×9106 =1810−910 =910=0.9Ns