a crane drops a 1.0 kg steel ball onto a steel plate. the ball's speed just before impact is 2.0 m/s. if the average impact force is 300 N, and the ball is in contact with the plate for 0.01 seconds, what is the bounce back speed right after the impact.
F=Δp/Δt
FΔt=Δp =p₂-(-p₁)=mv₁+mv₂
(since p₁ directed ↓and p₂ ↑)
v₂ = {FΔt- mv₁}/m =
={300•0.01 - 1•2}/1 = 1 m/s
To find the bounce back speed of the steel ball after the impact, we need to use the principle of conservation of momentum.
1. Start by calculating the initial momentum of the ball before the impact. The momentum (p) of an object is given by the product of its mass (m) and its velocity (v). In this case, the mass of the ball is 1.0 kg, and the velocity is 2.0 m/s. Therefore, the initial momentum (p_i) is equal to 1.0 kg × 2.0 m/s = 2.0 kg·m/s.
2. According to the principle of conservation of momentum, the total momentum before the impact is equal to the total momentum after the impact. Since the ball is the only object involved, its momentum is conserved. Therefore, the final momentum (p_f) of the ball after the impact will also be 2.0 kg·m/s.
3. Next, we need to calculate the change in momentum (Δp) during the impact. The change in momentum is given by the difference between the final and initial momentum: Δp = p_f - p_i. In this case, Δp = 2.0 kg·m/s - 2.0 kg·m/s = 0 kg·m/s.
4. Now, we can calculate the average force (F_avg) exerted on the ball during the impact using the formula: F_avg = Δp / Δt, where Δt is the time during which the ball is in contact with the plate. In this case, Δt = 0.01 seconds and F_avg = 300 N. Rearranging the formula, we get Δp = F_avg × Δt = 300 N × 0.01 s = 3.0 kg·m/s.
5. We know that the change in momentum is equal to the impulse (J) experienced by the ball, which is given by the product of the average force and the contact time: Δp = J = F_avg × Δt.
6. Finally, we can calculate the bounce back speed (v_f) using the formula: v_f = (Δp + p_i) / m, where m is the mass of the ball. In this case, m = 1.0 kg. Plugging in the values, we get v_f = (0 kg·m/s + 2.0 kg·m/s) / 1.0 kg = 2.0 m/s.
Therefore, the bounce back speed of the steel ball right after the impact is 2.0 m/s.