Question

A bullet of mass 20gm and with the velocity of 150m/sec moving horizontally strikes a

wooden material and comes to rest in 0.02sec.Calculate the magnitude of force exerted by

the wooden material on the bullet.


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Answer 1

Answer:

Mass of bullet (m)= 20 g = 0.02 kg

Initial velocity of bullet (u) =100 m/s

Final velocity of bullet (V) = 0

Time (t)= 0.05 seconds

Acceleration on the bullet (a)?

Force acting on wooden block (F) = ?

Distance penetrated by bullet (s) = ?

We know;

Step 1: Calculate the acceleration of the v = u + at

0= 100+ a ×0.05

– a × 0.05 = 100

a = -2000 ms-2

Step 2: Calculate the distance of penetration using s = ut + ½ at2

= 100 × 0.05 + 0.5 x (–2000) × (0.05)2

= 7.5 m

We know, F = ma

Force acting on bullet (F)

= 0.02 × (– 2000) = -40N

Negative sign denotes that wooden block exerts force in the direction, opposite to the direction of motion of the bullet.

Explanation:

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