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Q.

A bullet of mass 10 g travelling horizontally with a velocity of 150 m/s strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the magnitude of the force exerted by the wooden block on the bullet.


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a

45 N

b

-50 N

c

35 N

d

-40 N 

answer is B.

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Detailed Solution

The magnitude of the force exerted by the wooden block on the bullet is -50 N.
Given,
Mass of bullet, = 10 g .
Or,     m=0.01 kg
Initial velocity, u=150 m/s
Final velocity, v=0   [as it comes to rest finally]
Time, t=0.03 s
Now, Acceleration, a= - ut           =0 - 150 0.03  
                       = -5000 m/s2
We know that, = m×a   = 0.01 × ( - 5000) 
   F= -50 N
A negative sign indicates that the force is acting in the opposite direction of the motion of the bullet.
 
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