Question

A cricket ball of mass 150 g is moving with a velocity of 12 m/s and is hit by a bat so that the ball is turned back with a velocity of 20 m/s. If duration of contact between the ball and the bat is 0.01 sec. The impulse of the force is 

(A) 7.4 NS 

(B) 4.8 NS 

(C) 1.2 NS 

(D) 4.7 NS

Answer 1

Correct option: (B) 4.8 NS

Explanation:

m₁ = 0.15 kg

v₁ = 12

v₂ = 20

t = 0.01

F × Δt = change in momentum

= m₁v₁ – m₂v₂ 

here ball is turned back

F × Δt = m₁v₁ + m₂ v₂ 

= m(v₁ + v₂) --- mass is same

F × Δt = 0.15(12 + 20)

= 4.8

impulse of force = F × Δt = 4.8 NS