A 2.1-kg brick Determine the is placed gently upon a 2.9-kg cart originally movi
ID: 1425319 • Letter: A
Question
A 2.1-kg brick Determine the is placed gently upon a 2.9-kg cart originally moving with a speed of 26 cm/s. Determine the post-collision speed of the combination of brick and cart. A 98-kg fullback is running along at 8.6 m/s when a 76-kg defensive back running in the same direction at 9.8 m/s jumps on his back. What is the post-collision speed of the two players immediately after the tackle? A 0.112-kg billiard ball moving at 154 cm/s strikes a second billiard ball of the same mass moving in the opposite direction at 46 cm/s. The second billiard ball rebounds and travels at 72 cm/s after the head-on collision. Determine the post-collision velocity of the first billiard ball. A 225-kg bumper car (and its occupant) is moving north at 98 cm/s when it hits a 198-kg car (occupant mass included) moving north at 28 cm/s. The 198-kg car is moving north at 71 cm/s after the head-on collision. Determine the post-collision velocity of the 225-kg car. A 4.88-kg bowling ball moving east at 2.41 m/s strikes a stationary 0.95-kg bowling pin. Immediately after the head-on collision, the pin is moving east at 5.19 m/s. Determine the post-collision velocity of the bowling ball.Explanation / Answer
2)
let the final velocity is v
Using conseravtion of momentum
(2.1 + 2.9) * v = 2.9 * 26
solving for v
v = 15.1 cm/s
the final velocity of the system is 15.1 cm/s
3)
let the final velocity is v
Using conseravtion of momentum
(89 + 76) * v = 98 * 8.6 - 76 * 9.8
solving for v
v = 0.59 m/s
the post collision velocity of the system is 0.59 m/s
4)
let the final speed of ball is v m/s
intial sum of momentum = final sum of momentum
0.112 * 154 - 0.112 * 46 = 0.112 * v + 0.112 * 72
solving for v
v = 36 cm/s
the post collision velocity of first ball is 36 cm/s
5)
let the post velocity velocity is v
Using conservation of momentum
225 * 98 + 198 * 28 = 198 * 71 + 225 * v
solving for v
v = 60.16 cm/s
the post collision velocity of 225 Kg is 60.16 cm/s
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.