Two billiard balls of mass 0.05 kg each moving in opposite directions with 10 ms -1 collide and rebound with the same speed. If the time duration of contact is t =0.005 s, then what is the force exerted on the ball due to each other?

Question

Two billiard balls of mass 0.05 kg each moving in opposite directions with 10 ms -1 collide and rebound with the same speed. If the time duration of contact is t =0.005 s, then what is the force exerted on the ball due to each other? (A) 100 N (B) 200 N (C) 300 N (D) 400 N

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/d051b0b5565618e0d5d0a10d37ec454b-.png" /> Change in momentum of any one ball |Δ P |=2 × 0.05 × 10 |Δ P |=1 | F av |=(|Δ P |/Δ t ) F av =200 N

Q. Two billiard balls of mass $0.05 k g$ each moving in opposite directions with $10 m s^{- 1}$ collide and rebound with the same speed. If the time duration of contact is $t = 0.005 s$ , then what is the force exerted on the ball due to each other?

100 N

200 N

300 N

400 N

Change in momentum of any one ball
$∣Δ P ∣ = 2 \times 0.05 \times 10$
$∣Δ P ∣ = 1$
$∣ ∣ F_{a v} ∣ ∣ = \frac{∣Δ P ∣}{Δ t}$
$F_{a v} = 200 N$

Q. Two billiard balls of mass 0.05kg each moving in opposite directions with 10ms−1 collide and rebound with the same speed. If the time duration of contact is t=0.005s, then what is the force exerted on the ball due to each other?