The Student Room Group

This discussion is no longer active so you won't be able to reply.Check out other Related discussions

A2 physics inelastic collisions question

I keep on getting a higher value for kinetic energy after the collision in the question which seems wrong and i really cant see why i keep on getting this problem. Any help would be appreciated.

Q. An ice puck of mass 1.5kg moving at 4.2ms^-1 collides head on with a second puck of mass 1.0kg moving in the opposite direction at 4.0 ms^-1. After impact the 1.5kg puck continues in the same direction at 0.8ms^-1. Calculate
a) the speed and direction of the 1.0kg puck after the collision
b) the loss of kinetic energy due to the collision.

A.
a) ( (1.5 X 4.2) + (1 X 4) ) - 1.2 = 9.1ms^-1
b) 1/2(1)(4)^2 +1/2(1.5)(4.2)^2 = 21.23J
1/2(1)(9.1)^2 + 1/2(1.5)(0.8)^2 = 41.885J

Somehow the KE has increased so none is lost which must be wrong.
Original post by medleyboi
I keep on getting a higher value for kinetic energy after the collision in the question which seems wrong and i really cant see why i keep on getting this problem. Any help would be appreciated.

Q. An ice puck of mass 1.5kg moving at 4.2ms^-1 collides head on with a second puck of mass 1.0kg moving in the opposite direction at 4.0 ms^-1. After impact the 1.5kg puck continues in the same direction at 0.8ms^-1. Calculate
a) the speed and direction of the 1.0kg puck after the collision
b) the loss of kinetic energy due to the collision.

A.
a) ( (1.5 X 4.2) + (1 X 4) ) - 1.2 = 9.1ms^-1
b) 1/2(1)(4)^2 +1/2(1.5)(4.2)^2 = 21.23J
1/2(1)(9.1)^2 + 1/2(1.5)(0.8)^2 = 41.885J

Somehow the KE has increased so none is lost which must be wrong.


The 'hidden task' in this question is that you need to convert speed to velocity.

Momentum is a vector so when working out a collision between objects going in *opposite* directions like this you have to set the velocity of one of the objects to be -ve before you attempt your conservation of momentum calculations.

fwiw KE is scalar - a mass travelling at velocity v has the same KE as a mass travelling at velocity -v as you should be able to confirm easily by plugging -ve and +ve values of v into 1/2 mv2... so you can be consistent in your use of -ve velocity throughout.
Reply 2
A)

Original post by medleyboi
I keep on getting a higher value for kinetic energy after the collision in the question which seems wrong and i really cant see why i keep on getting this problem. Any help would be appreciated.

Q. An ice puck of mass 1.5kg moving at 4.2ms^-1 collides head on with a second puck of mass 1.0kg moving in the opposite direction at 4.0 ms^-1. After impact the 1.5kg puck continues in the same direction at 0.8ms^-1. Calculate
a) the speed and direction of the 1.0kg puck after the collision
b) the loss of kinetic energy due to the collision.

A.
a) ( (1.5 X 4.2) + (1 X 4) ) - 1.2 = 9.1ms^-1
b) 1/2(1)(4)^2 +1/2(1.5)(4.2)^2 = 21.23J
1/2(1)(9.1)^2 + 1/2(1.5)(0.8)^2 = 41.885J

Somehow the KE has increased so none is lost which must be wrong.

Your question A is wrong it should be ( (1.5 X 4.2) + (1 X -4) ) - 1.2 = 1.1ms^-1 since its travelling in the oppsite direction so it should have negative velocity.

Latest