1- A 4.3-kg object moves to the right with a speed of 5.0 m/s.It collides in a p

Question

1- A 4.3-kg object moves to the right with a speed of 5.0 m/s.It collides in a perfectly elastic collision with a 5.9-kgobject moving to the left at 2.7 m/s. What is the total kineticenergy after the collision?
Total KE after collision? 2-A 63-kg astronaut is space-walking outside the space capsuleand is stationary when the tether line breaks. As a means ofreturning to the capsule he throws his 2.0-kg space hammer at aspeed of 17 m/s away from the capsule. At what speed does theastronaut move toward the capsule?
Speed of astronaut in m/s 3-What is the momentum of a 1300-kg truck traveling at 27m/s?
Truck's momentum in kg m/s 1- A 4.3-kg object moves to the right with a speed of 5.0 m/s.It collides in a perfectly elastic collision with a 5.9-kgobject moving to the left at 2.7 m/s. What is the total kineticenergy after the collision?
Total KE after collision? 1- A 4.3-kg object moves to the right with a speed of 5.0 m/s.It collides in a perfectly elastic collision with a 5.9-kgobject moving to the left at 2.7 m/s. What is the total kineticenergy after the collision?
Total KE after collision? 2-A 63-kg astronaut is space-walking outside the space capsuleand is stationary when the tether line breaks. As a means ofreturning to the capsule he throws his 2.0-kg space hammer at aspeed of 17 m/s away from the capsule. At what speed does theastronaut move toward the capsule?
Speed of astronaut in m/s 3-What is the momentum of a 1300-kg truck traveling at 27m/s?
Truck's momentum in kg m/s 3-What is the momentum of a 1300-kg truck traveling at 27m/s?
Truck's momentum in kg m/s 3-What is the momentum of a 1300-kg truck traveling at 27m/s?
Truck's momentum in kg m/s

Explanation / Answer

Mass m = 4.3 kg

          M = 5.9kg

Initial velocities u = 5 m / s

                          U = - 2.7 m / s

For perfectly elastic collision coeffient of restitution e =1

           (V – v) / ( u – U ) = 1

                             V – v = 7.7  

                               V = v + 7.7   ---( 1)

From law of conservation of momentum , mu + MU = mv + MV

              21.5 – 15.93 = 4.3 v +5.9 V

4.3 v +5.9 V = 5.57

From eq ( 1 ) , 4.3 v + 5.9 ( v + 7.7 ) = 5.57

                        8.2 v + 45.43 = 5.57

                                 v = -4.86 m / s

So, V = -4.86 + 7.7 = 2.839 m /s

Total kinetic energy after colliiosn K = ( 1/ 2) mv ^ 2+ (½ ) M V ^ 2

           K = 50.78 + 23.77 = 74.55 J

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