<< Chapter < Page | Chapter >> Page > |
A 0.240-kg billiard ball that is moving at 3.00 m/s strikes the bumper of a pool table and bounces straight back at 2.40 m/s (80% of its original speed). The collision lasts 0.0150 s. (a) Calculate the average force exerted on the ball by the bumper. (b) How much kinetic energy in joules is lost during the collision? (c) What percent of the original energy is left?
(a) 86.4 N perpendicularly away from the bumper
(b) 0.389 J
(c) 64.0%
During an ice show, a 60.0-kg skater leaps into the air and is caught by an initially stationary 75.0-kg skater. (a) What is their final velocity assuming negligible friction and that the 60.0-kg skater’s original horizontal velocity is 4.00 m/s? (b) How much kinetic energy is lost?
Professional Application
Using mass and speed data from [link] and assuming that the football player catches the ball with his feet off the ground with both of them moving horizontally, calculate: (a) the final velocity if the ball and player are going in the same direction and (b) the loss of kinetic energy in this case. (c) Repeat parts (a) and (b) for the situation in which the ball and the player are going in opposite directions. Might the loss of kinetic energy be related to how much it hurts to catch the pass?
(a) 8.06 m/s
(b) -56.0 J
(c)(i) 7.88 m/s; (ii) -223 J
A battleship that is $6\text{.}\text{00}\times {\text{10}}^{7}\phantom{\rule{0.25em}{0ex}}\text{kg}$ and is originally at rest fires a 1100-kg artillery shell horizontally with a velocity of 575 m/s. (a) If the shell is fired straight aft (toward the rear of the ship), there will be negligible friction opposing the ship’s recoil. Calculate its recoil velocity. (b) Calculate the increase in internal kinetic energy (that is, for the ship and the shell). This energy is less than the energy released by the gun powder—significant heat transfer occurs.
Professional Application
Two manned satellites approaching one another, at a relative speed of 0.250 m/s, intending to dock. The first has a mass of $4\text{.}\text{00}\times {\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{kg}$ , and the second a mass of $\text{7.50}\times {\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{kg}$ . (a) Calculate the final velocity (after docking) by using the frame of reference in which the first satellite was originally at rest. (b) What is the loss of kinetic energy in this inelastic collision? (c) Repeat both parts by using the frame of reference in which the second satellite was originally at rest. Explain why the change in velocity is different in the two frames, whereas the change in kinetic energy is the same in both.
(a) 0.163 m/s in the direction of motion of the more massive satellite
(b) 81.6 J
(c) $8\text{.}\text{70}\times {\text{10}}^{-2}\phantom{\rule{0.25em}{0ex}}\text{m/s}$ in the direction of motion of the less massive satellite, 81.5 J. Because there are no external forces, the velocity of the center of mass of the two-satellite system is unchanged by the collision. The two velocities calculated above are the velocity of the center of mass in each of the two different individual reference frames. The loss in KE is the same in both reference frames because the KE lost to internal forces (heat, friction, etc.) is the same regardless of the coordinate system chosen.
Notification Switch
Would you like to follow the 'College physics' conversation and receive update notifications?