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# GCSE momentum Watch

1. Hi so i know how to work out nmomentum using the original formula (momentum=mass x velocity) and i can re arrange
but im stuck on working out momentum after a collision and velocity after a collision
etc

A 0.5kg trolley A is pushed at a velocity of 1.2m/s into a stationary trolley B of mass 1.5kg. The two trolleys stick to each other after the impact. What is the momentum of trolley A before the collision. What is the velocity of the two trolleys after the collision?

can anyone help?
Also how do you use the conservation of momentum when working out using equations/formulas?
thanks
2. A 0.5kg trolley A is pushed at a velocity of 1.2m/s into a stationary trolley B of mass 1.5kg. The two trolleys stick to each other after the impact. What is the momentum of trolley A before the collision. What is the velocity of the two trolleys after the collision?
Total momentum before=0.5*1.2+1.5*0=0.6 kgm/s
Total momentum after=total energy before (by conservation of linear momentum)
So Total momentum after= 0.6 =(0.5+1.5)*v
So v=0.6/2=0.3 m/s
3. I got 0.3m/s as my answer
4. (Original post by XShmalX)
A 0.5kg trolley A is pushed at a velocity of 1.2m/s into a stationary trolley B of mass 1.5kg. The two trolleys stick to each other after the impact. What is the momentum of trolley A before the collision. What is the velocity of the two trolleys after the collision?
Total energy before=0.5*1.2+1.5*0=0.6 kgm/s
Total energy after=total energy before (by conservation of linear momentum)
So Total energy after= 0.6 =(0.5+1.5)*v
So v=0.6/2=0.3 m/s
All of the maths here is correct, and has lead to the right answer. Do note though that we are talking about momentum rather than energy; they are quite different. Replace every instance of "energy" with "momentum" and it's perfect!
5. (Original post by halii_94)
Hi so i know how to work out nmomentum using the original formula (momentum=mass x velocity) and i can re arrange
but im stuck on working out momentum after a collision and velocity after a collision
etc

A 0.5kg trolley A is pushed at a velocity of 1.2m/s into a stationary trolley B of mass 1.5kg. The two trolleys stick to each other after the impact. What is the momentum of trolley A before the collision. What is the velocity of the two trolleys after the collision?

can anyone help?
Also how do you use the conservation of momentum when working out using equations/formulas?
thanks
For future reference, when the trolleys stick to each other you can take their masses as one mass.
6. (Original post by Pangol)
All of the maths here is correct, and has lead to the right answer. Do note though that we are talking about momentum rather than energy; they are quite different. Replace every instance of "energy" with "momentum" and it's perfect!
doh! Having a totally brainstorm!
7. (Original post by XShmalX)
A 0.5kg trolley A is pushed at a velocity of 1.2m/s into a stationary trolley B of mass 1.5kg. The two trolleys stick to each other after the impact. What is the momentum of trolley A before the collision. What is the velocity of the two trolleys after the collision?
Total momentum before=0.5*1.2+1.5*0=0.6 kgm/s
Total momentum after=total energy before (by conservation of linear momentum)
So Total energy after= 0.6 =(0.5+1.5)*v
So v=0.6/2=0.3 m/s
why do you divide it by 2 if the velocity is 1.2m/s?
8. (Original post by halii_94)
why do you divide it by 2 if the velocity is 1.2m/s?
The total initial momentum is calculated by multiplying the mass of each trolley by its velocity.
The momentum before must equal the total momentum after by conservation of momentum.
So as the two trolleys stick together the momentum is divided by the totally mass of the two trolleys stuck together in order to find the velocity.
9. (Original post by halii_94)
why do you divide it by 2 if the velocity is 1.2m/s?
Momentum of A = mass x velocity
= 0.5 x 1.2
= 0.6kgm/s (momentum)

Momentum before collision = momentum after the collision

0.6 = mass of the two trolleys combined (says they stick) x velocity of trollies.

0.6 = 2 x Velocity

Velocity = 0.3m/s
10. (Original post by XShmalX)
The total initial momentum is calculated by multiplying the mass of each trolley by its velocity.
The momentum before must equal the total momentum after by conservation of momentum.
So as the two trolleys stick together the momentum is divided by the totally mass of the two trolleys stuck together in order to find the velocity.
so if there were 3 etc trolleys it would divide by 3?
11. (Original post by gozatron)
Momentum of A = mass x velocity
= 0.5 x 1.2
= 0.6kgm/s (momentum)

Momentum before collision = momentum after the collision

0.6 = mass of the two trolleys combined (says they stick) x velocity of trollies.

0.6 = 2 x Velocity

Velocity = 0.3m/s
thank you
12. (Original post by halii_94)
so if there were 3 etc trolleys it would divide by 3?
No you'd divide by the sum of the masses of the three trolleys. It just happens that in this case the two mass are 0.5 kg and 1.5 kg and so equal 2 kg.
13. (Original post by XShmalX)
No you'd divide by the sum of the masses of the three trolleys. It just happens that in this case the two mass are 0.5 kg and 1.5 kg and so equal 2 kg.
thank youu
14. (Original post by XShmalX)
No you'd divide by the sum of the masses of the three trolleys. It just happens that in this case the two mass are 0.5 kg and 1.5 kg and so equal 2 kg.
one last thing

A car of mass 950kg is travelling at 20m/s when it shunts a car of mass 1000kg which is travelling at 15m/s
what is the velocity of the two cars after the collision?

i got the answer 17.44 m/s
does that sound right?
thanks a lot
15. (Original post by halii_94)
one last thing

A car of mass 950kg is travelling at 20m/s when it shunts a car of mass 1000kg which is travelling at 15m/s
what is the velocity of the two cars after the collision?

i got the answer 17.44 m/s
does that sound right?
thanks a lot
Assuming they are traveling at the same speed, but otherwise the question wouldn't be possible, so yes!
16. (Original post by XShmalX)
Assuming they are traveling at the same speed, but otherwise the question wouldn't be possible, so yes!
yay thanks you it has taken me months literally to understand momentum lol

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