Momentum questions - A-Level onwards standard (formulas and explanations needed)

[tomclaessens]

Homework Statement

I am a GCSE Physics student, and my teacher recently set us a homework sheet in which he said contained questions 'exceeding A-Level standard'. He told us to get as far as we could. So, I have completed 12 of 18 questions, but I am very confused about the last 6:

1) A 3kg trolley moving at 40cm/s collides with and joins to a stationary trolley of mass 2kg on a horizontal, frictionless surface. Calculate the speed at which the trolleys move after the collision and the loss of kinetic energy.

2) Two trolleys, each of mass 2kg, collide and stick together on a smooth, horizontal surface. One trolley is at rest before the collision. The other moves at 20cm/s before the collision. Calculate the combined velocity (size and direction) of the trolleys after the collision.

3) Two trolleys, both of mass 2kg, are moving in the same direction along a smooth surface. One is moving faster (30cm/s) and catches up the other (20cm/s). The trolleys collide and stick together. Calculate the combined velocity (size and direction) of the trolleys after the collision.

4) Two trolleys (one of mass 2kg, traveling at 30cm/s and one of mass 3kg, traveling at 20cm/s) are moving in opposite directions along a smooth surface. The trolleys collide and stick together.
a) What is the total momentum of the trolleys before and after the collision?
b) What is the trolleys' combined velocity after the collision?

5) A car of mass 1500kg, traveling at 15m/s, collides 'head on' with a second car, mass 1000kg, which has a speed of 16.5m/s in the opposite direction. Find the speed and direction of the cars immediately after the collision, assuming that they remain locked together.

6) Two trolleys, of mass 2kg and 5kg, are at rest and in contact on a smooth, level surface. A coiled spring in one trolley is released so that they 'explode' apart. The lighter trolley moves off at 50cm/s. Find the speed of the heavier trolley and the minimum energy which was stored in the coiled spring before release.

Homework Equations

I am unsure of the necessary equations to complete these questions, and would appreciate anyone who could tell me them, what they mean and how I can apply them to work out the answers.

The Attempt at a Solution

I have tried for 3 days to find a solution to these questions, searching for formulas and asking past A-Level Physics students, but I am really struggling to understand how to work the answers out.

Many thanks,

tomclaessens

 

[Doc Al]

Hint: All six involve conservation of momentum. In the first five, that's all you need.

Give them a shot.

 

[tomclaessens]

I know they involve the convservation of momentum - that the initial momentum has to be the same as the momentum after the collision, but I don't understand how to use that to work it out?

Believe me, I have given it a shot...

 

[Doc Al]

What's the equation for conservation of momentum during a collision?

 

[tomclaessens]

I didn't know, but I found this on this internet:

'm1 u1 + m2 u2 = (m1 + m2) v'
Is this correct and if so, what does it mean please?

 

[Boogeyman]

I didn't know, but I found this on this internet:

'm1 u1 + m2 u2 = (m1 + m2) v'
Is this correct and if so, what does it mean please?

It means the initial momentum is equal to the final momentum. You just need to find the total momentum before the collision and the total momentum after. V is the common velocity at which both bodies move after the collision.

 

[tomclaessens]

Thank you, but how can I find the momentum after the collision if I am not given the velocity? (For example, in question 1: it is the velocity I need to calculate (as well as the loss of kinetic energy))

 

[tomclaessens]

For question number 1, I have attempted a solution again. Please could someone tell me whether this is correct or not:

3kg x 40cm/s = 1.2kgm/s
therefore the total momentum after the collision is also 1.2kgm/s
to calculate the speed, I divided 1.2kgm/s by 2kg, which gave me 0.6m/s or 60cm/s as the speed at which the trolleys travel at after the collision

 

[Doc Al]

Thank you, but how can I find the momentum after the collision if I am not given the velocity? (For example, in question 1: it is the velocity I need to calculate (as well as the loss of kinetic energy))

You're supposed to solve for that velocity. Just set up the equation and solve for the only unknown--the final speed.

For question number 1, I have attempted a solution again. Please could someone tell me whether this is correct or not:

3kg x 40cm/s = 1.2kgm/s
therefore the total momentum after the collision is also 1.2kgm/s

OK.

to calculate the speed, I divided 1.2kgm/s by 2kg, which gave me 0.6m/s or 60cm/s as the speed at which the trolleys travel at after the collision

Why did you divide by 2 kg? Compare what you did to the equation you wrote in post #5.

Also, take a look at this: https://www.physicsforums.com/showpost.php?p=2157983&postcount=5"

 

[tomclaessens]

Thanks for everyone's help.
It just needed someone to explain it to me, which my friend (a past A-Level physics student) kindly did.

tomclaessens