Question

A bullet of mass 15 g travelling horizontally with a velocity of 720kmh-1 strikes a stationary wooden block and comes to rest in 0.06 sec. The magnitude of the force exerted by the wooden block on the bullet is
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Answer 1

Concept

• The bullet is moving with velocity and is retarded uniformly and finally comes at rest.
• This retardation is provided by the resistive force exerted by wooden block.

》v² = u² + 2as

》F = ma

Solution

Given

v = final velocity = 0(rest)

u = initial velocity = 720 km/h = 200 m/s

t = 0.06 s

m = 15g = 0.015 kg

a = ?

• v = u + at
• 0 = 200 + (-a)(0.06)
• 200 = 0.06 a
• a = 10⁴ /3

F = ma = (0.015 × 10⁴)/3

= 50 N

★ Resistive force exerted is 50N

Answer 2

Explanation:

Concept

• The bullet is moving with velocity and is retarded uniformly and finally comes at rest.
• This retardation is provided by the resistive force exerted by wooden block.

》v² = u² + 2as

》F = ma

Solution

Given

v = final velocity = 0(rest)

u = initial velocity = 720 km/h = 200 m/s

t = 0.06 s

m = 15g = 0.015 kg

a = ?

v = u + at

0 = 200 + (-a)(0.06)

200 = 0.06 a

a = 10⁴ /3

F = ma = (0.015 × 10⁴)/3

= 50 N

★ Resistive force exerted is 50N