Question

Q9.A bullet of mass 0.04 kg moving with a speed of 90 m/s enters a heavy wooden block and stopped after 3s. What is the average resistive force exerted by the block on the bullet? A. 1N B. 2N C. 3N D. 1.2N
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Answer 1

Given :-

• Mass of bullet is(m= 0.04 kg )
• Moving with speed(u= 90 m/s )
• It is stopped i.e Final velocity (v=0 m/s)
• Time taken is (t = 3s)

To find :-

• Force exerted by block on bullet

Explanation:-

As we need to find the force acting on the body. We can find using Newton's second law of equation.

F = m(v-u)/t

Substituting the values,

F= 0.04(0-90)/3

F = 0.04(-90)/3

F = 0.04 (-30)

F = 4×10^-2 × -3× 10

F = -12×10^-1

F = -12/10

F = -1.2 N

So, force exerted by the block on bullet is 1.2N (in opposite direction) .

Thus,

The correct answer is option-D (1.2N)