On a frictionless, horizontal air table, puck A (with mass0.250 kg) is moving to
ID: 1722959 • Letter: O
Question
On a frictionless, horizontal air table, puck A (with mass0.250 kg) is moving toward puck B (with mass 0.350 kg), which is initially at rest. After thecollision, puck A has a velocity of 0.120 m/s to the left,and puck B has a velocity of 0.660 m/s to the right. (a) What was the speed of puck Abefore the collision?1 m/s
(b) Calculate the change in the total kinetic energy of the systemthat occurs during the collision.
2 J (a) What was the speed of puck Abefore the collision?
1 m/s
(b) Calculate the change in the total kinetic energy of the systemthat occurs during the collision.
2 J
Explanation / Answer
Let the speed towards left be -ve and speedstowards right be + ve. a. Total lnear momentumbefore collision = total linearmomentum after collision mA *uA + mB *uB = mA *vA + mB *vB m is mass, u is velocitybefore collision and v is velocity after collision. mA = 0.250 kg, mB = 0.350 kg, vA = -0.120 m/s, vB = 0.660 m/s, uB = 0 => 0.250 *uA + 0.350 *0 = 0.250 * ( -0.120) + 0.350 * 0.660 Velocity of puck A beforecollision uA = 0.201/0.250 = 0.804 m/s + ve sign indicates motion towardsright. b. Total initialk.e. KI = (1/2)* mA *uA2 + (1/2)* mB * uB2 = 0.5* 0.250 * 0.8042 + 0.5* 0.350 * 02 = 8.08 *10-2 J Totalfinal k.e. KF = (1/2)* mA *vA2 + (1/2)* mB * vB2 = 0.5* 0.250 * ( -0.120)2 + 0.5 * 0.350* 0.6602 = 7.80 *10-2 J Change in k.e. K = KF - KI = 7.80* 10-2 - 8.08 *10-2 = - 2.80 *10-3 J - ve sign indicates loss of k.e. = 0.5* 0.250 * ( -0.120)2 + 0.5 * 0.350* 0.6602 = 7.80 *10-2 J Change in k.e. K = KF - KI = 7.80* 10-2 - 8.08 *10-2 = - 2.80 *10-3 J - ve sign indicates loss of k.e.Related Questions
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