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Questions about momentum 2D

1) One mass of 5.8x10^6 kg collides with an asteroid in space with a mass of 1.2x10^12kg.


The two bodies join together in impact, the speed of the colliding mass was taken as 35,000 ms^1 and the asteroid was 25,000 ms^1.

Is this an elastic or inelastic collision?

2) Hence calculate the resulting change in the asteroid's direction

3) Such an impact may fracture the asteroid and leave some fragments still on a collision course with the Earth. An alertnative approach would be to apply a smaller steady force over a longer period, for example by using rocket engines.

State a formula for net force in terms of momentum.

4) The momentum change of Castalia at the impact if 2x10^11 Ns. The average thrust of a typical rocket is 7x10^5N, and it can burn for 130s. Calculate the number of rockets required to produce the same change in Castalia's momentum as in the collision above.
Original post by The Best1
1) One mass of 5.8x10^6 kg collides with an asteroid in space with a mass of 1.2x10^12kg.


The two bodies join together in impact, the speed of the colliding mass was taken as 35,000 ms^1 and the asteroid was 25,000 ms^1.

Is this an elastic or inelastic collision?

2) Hence calculate the resulting change in the asteroid's direction

3) Such an impact may fracture the asteroid and leave some fragments still on a collision course with the Earth. An alertnative approach would be to apply a smaller steady force over a longer period, for example by using rocket engines.

State a formula for net force in terms of momentum.

4) The momentum change of Castalia at the impact if 2x10^11 Ns. The average thrust of a typical rocket is 7x10^5N, and it can burn for 130s. Calculate the number of rockets required to produce the same change in Castalia's momentum as in the collision above.


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Reply 2
1) I wrote it as inelastic collision because inelastic collisions have both the colliding objects as one piece (or at least I think). I know in elastic collisions, the kinetic energy of both the colliding molecule is not converted into sound energy etc. but inelastic does have it converted into sound energy, so some energy is lost. Overall, the rule is that momentum is conserved.

What makes me think it might be elastic is the fact that it is space, and if there are no particles, energy cannot be converted (I think?)..

2) It also says that the mass strikes Castalia at 90 degrees, so I drew a triangle with the two right angle sides repressenting the momentum (mass x velocity) each. The hypotenuse line I found by using a^2 + b^2 all square rooted, which the value was 2.7x10^8 Kgms^-1.

I cannot find the 'resulting change' in the direction... because when i use SOH CAH or TOA in this triangle, it says invalid answer for the angle..

3) I wrote my answer as total momentum (mv) = (1)mv + (2) mv

4) Had no idea..
Reply 3
NVM for question 2, I've realised the value for the hypotenuse line is actually 5.16x10^16. Therefore SOH CAH TOA does work, so I get the angle as 54.45 degrees
Original post by The Best1
NVM for question 2, I've realised the value for the hypotenuse line is actually 5.16x10^16. Therefore SOH CAH TOA does work, so I get the angle as 54.45 degrees


Yeah inelastic, the 'lost' energy from a collision doesn't need to be sound, it could be converted to internal energy i.e. heat up the asteroid.

fwiw there has been a man made high speed collision between a spacecraft and a comet (deep impact/comet tempel 1), there's video of it on youtube and it's pretty clear that a lot of heat is developed and radiated away.

for part 4 you can calculate the momentum change produced per rocket from the given thrust per rocket engine (units is N) and duration per rocket engine (unit is s)... then calculate the number of rockets required to reach the target change in momentum (unit is Ns)

It's an important point that impulse:
force x time (units Ns) from a rocket engine or a collision
is the exact same thing in the same units as a change in momentum:
mass x velocity (units Kg ms^-1)... examiners might try throwing you off with this.

see here http://www.bbc.co.uk/education/guides/zv3j6sg/revision/3
Original post by The Best1
NVM for question 2, I've realised the value for the hypotenuse line is actually 5.16x10^16. Therefore SOH CAH TOA does work, so I get the angle as 54.45 degrees


hey, could you explain how you got that? I keep getting 3 x 10^16 as the hypotenuse

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