Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

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

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote