a 1.0 kg ball is thrown horizontally with a velocity 15 m/s against a wall. If the ball rebounds horizontally with a velocity of 13 m/s and the contact time is 0.020s what is the force exerted on the ball by the wall?
1400N
To find the force exerted on the ball by the wall, we can use Newton's second law of motion, which states that force (F) is equal to the rate of change of momentum (mv) over time (t).
The initial momentum of the ball before hitting the wall is given by the mass (m) multiplied by the initial velocity (u). Therefore, the initial momentum (p₁) is equal to 1.0 kg multiplied by 15 m/s, giving us:
p₁ = 1.0 kg * 15 m/s = 15 kg·m/s
The final momentum of the ball after rebounding is given by the mass (m) multiplied by the final velocity (v). Therefore, the final momentum (p₂) is equal to 1.0 kg multiplied by 13 m/s, giving us:
p₂ = 1.0 kg * 13 m/s = 13 kg·m/s
Since momentum is conserved in this case (no external forces acting on the ball horizontally), the change in momentum (Δp) is equal to the negative of the initial momentum:
Δp = p₂ - p₁ = 13 kg·m/s - 15 kg·m/s = -2 kg·m/s
Now, we can find the force (F) exerted on the ball by dividing the change in momentum (Δp) by the contact time (t):
F = Δp / t = (-2 kg·m/s) / (0.020 s) = -100 N
Note that the force here is negative, indicating that it is in the opposite direction (since the ball rebounds horizontally). Therefore, the force exerted on the ball by the wall is 100 N in the opposite direction of the ball's motion.