A .060 gram handball is thrown straight toward a wall with aspeed of 10m/s. It r

Question

A .060 gram handball is thrown straight toward a wall with aspeed of 10m/s. It rebounds straight bakcward at a sped of8.0m/s. A.) What impulse is exerted on the wall? B.) If the ball is in contact with the wall for 3.0ms, whataverage force is exerted on the wall by the ball? C) The rebounding ball is caught by a player who brings it toa rest. In the process, her hand moves back .50m. What is theimpulse received by the player? D.) What average force was exerted on the player by theball? A .060 gram handball is thrown straight toward a wall with aspeed of 10m/s. It rebounds straight bakcward at a sped of8.0m/s. A.) What impulse is exerted on the wall? B.) If the ball is in contact with the wall for 3.0ms, whataverage force is exerted on the wall by the ball? C) The rebounding ball is caught by a player who brings it toa rest. In the process, her hand moves back .50m. What is theimpulse received by the player? D.) What average force was exerted on the player by theball?

Explanation / Answer

The mass of the handball is m = .060 gram = .060 *10-3 kg The handball is thrown straight toward a wall with aspeed of u = 10 m/s The handball rebounds straight bakcward at a speed of v = 8.0m/s A.)Let the time taken by the ball to rebound straight backwardbe t,therefore,we get v = u + gt Here,g = 9.8 m/s2 or t = (v - u/g) Here,v = -8.0 m/s and u = 10 m/s or t = (-8.0 - 10/9.8) = 1.8 s Therefore,the impulse that is exerted on the wall is I = F * t or I = m * g * t or I = .060 * 10-3 * 9.8 * 1.8 = 1.06 *10-3 N.s B.)When the ball is in contact with the wall for a time t =3.0 ms = 3.0 * 10-3 s,then the acceleration of the ballis a = (v - u/t) or a = (-8.0 - 10/3.0 * 10-3) = -6 * 103m/s2 The negative value of accelertion indicates that the motion ofthe ball is a deceleration. Therefore,the average force that is exerted on the wall by theball is F = m * a or F = .060 * 10-3 * 6 * 103 = 0.36N C.)Let the acceleration of the ball when it is caught by theplayer be a.Therefore,we get v2 - u2 = 2aS Here,v = 0 m/s,u = -8.0 m/s and S = .50 m or (0)2 - (-8.0)2 = 2a * .50 or a = -64 m/s2 Let the time taken by the ball to come to a rest bet,therefore,we get v = u + at or 0 = -8.0 + (-64)t or t = (8.0/64) = (1/8) = 0.125 s Therefore,the impulse received by the player is I = F * t = m * a * t or I = .060 * 10-3 * 64 * 0.125 or I = 0.48 * 10-3 N.s D.)The average force exerted on the player by the ballis F = m * a or F = .060 * 10-3 * 64 = 3.84 * 10-3N Let the time taken by the ball to come to a rest bet,therefore,we get v = u + at or 0 = -8.0 + (-64)t or t = (8.0/64) = (1/8) = 0.125 s Therefore,the impulse received by the player is I = F * t = m * a * t or I = .060 * 10-3 * 64 * 0.125 or I = 0.48 * 10-3 N.s D.)The average force exerted on the player by the ballis F = m * a or F = .060 * 10-3 * 64 = 3.84 * 10-3N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.