Hi, I've some mechanics questions to ask, those questions are from the OCR mechanics 1 book. Thanks.
1)Two particles A and B, with masses m1 and m2 respectively, move in the same straight line and collide. Immediately before the collision the particles are moving towards each other. A with speed 3 ms^1 and B with speed 2 ms^1. Immediately after the collision both particles are moving in the same direction, one with speed 3 ms^1 and the other with speed 2 ms^1.
b)Find the ratio m1:m2.
It is given that after the collision both particles are moving in the same direction, one with speed 3 ms^1 and the other with speed 2 ms^1. But how to know the direction is towards the left or to the right? And how to know which has speed of 3 or 2 ms^1?
I couldn't get the correct answer 5:1
2)A tennis ball of mass 60 grams is dropped on the floor from a height of 1.6 metres, and rebounds to a height of 1.2 metres. Calculate the impulse it receives when it hits the ground.
So v² = 2gh
Solving this gives v = 5.6 ms^1, then I don't really know what to do next. I've tried some ways but I get the wrong answer.
3)Three separate trucks, of mass 400 kg, 500 kg and 300 kg, are in line in that order on a railway track. Initially the second and third trucks are at rest, and the 400 kg truck is moving towards them at a speed of 12 ms^1. Each time that a moving truck hits a stationary truck at a speed of u ms^1, the difference in the speeds of the two trucks after the collision is 0.5 u ms^1. Find the speeds of the trucks after two collisions have taken place. Will there be any more collissions?
Firstly I don't understand that it says " Each time that a moving truck hits a stationary truck at a speed of u ms^1" while the moving truck is only the 400 kg truck and it is given a speed of 12 ms^1. I know other truck will be moving after collision but does that mean they will have the speed of u ms^1 and not other speed? I also don't understand why it gives us the information "the difference in the speeds of the two trucks after the collision is 0.5 u ms^1." So I am stuck.
Thanks for helping.

Civ217
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 06072006 16:06

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 06072006 16:25
(Original post by Civ217)
Hi, I've some mechanics questions to ask, those questions are from the OCR mechanics 1 book. Thanks.
1)Two particles A and B, with masses m1 and m2 respectively, move in the same straight line and collide. Immediately before the collision the particles are moving towards each other. A with speed 3 ms^1 and B with speed 2 ms^1. Immediately after the collision both particles are moving in the same direction, one with speed 3 ms^1 and the other with speed 2 ms^1.
b)Find the ratio m1:m2.
It is given that after the collision both particles are moving in the same direction, one with speed 3 ms^1 and the other with speed 2 ms^1. But how to know the direction is towards the left or to the right? And how to know which has speed of 3 or 2 ms^1?
I couldn't get the correct answer 5:1
2)A tennis ball of mass 60 grams is dropped on the floor from a height of 1.6 metres, and rebounds to a height of 1.2 metres. Calculate the impulse it receives when it hits the ground.
So v² = 2gh
Solving this gives v = 5.6 ms^1, then I don't really know what to do next. I've tried some ways but I get the wrong answer. 
silent ninja
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 06072006 16:49
(Original post by Civ217)
3)Three separate trucks, of mass 400 kg, 500 kg and 300 kg, are in line in that order on a railway track. Initially the second and third trucks are at rest, and the 400 kg truck is moving towards them at a speed of 12 ms^1. Each time that a moving truck hits a stationary truck at a speed of u ms^1, the difference in the speeds of the two trucks after the collision is 0.5 u ms^1. Find the speeds of the trucks after two collisions have taken place. Will there be any more collissions?
Firstly I don't understand that it says " Each time that a moving truck hits a stationary truck at a speed of u ms^1" while the moving truck is only the 400 kg truck and it is given a speed of 12 ms^1. I know other truck will be moving after collision but does that mean they will have the speed of u ms^1 and not other speed? I also don't understand why it gives us the information "the difference in the speeds of the two trucks after the collision is 0.5 u ms^1." So I am stuck.
Thanks for helping.
All it means is that after two trucks collide, the difference in speed of truck A with truck B, is 0.5u where u is the initial speed of A toward B.
You'll need to set up simultaneous equations because you dont know which direction or value of the speeds after collision. Conservation of momenum assuming they travel in the same direction after collision gives:
400*12 = 400*v1 + 500*v2
also we've been told that, v2v1= 0.5*12. solve.
do the same for the second truck colliding with third truck.
500*v2= 500*v3 + 300*v4
v4v3= 0.5v2
Whether there will be anymore collisions depends on which direction and speed the second truck travels after its collision with the third truck. If its velocity is less than the first truck, the first truck will collide with it again. 
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 06072006 17:14
(Original post by Civ217)
1)Two particles A and B, with masses m1 and m2 respectively, move in the same straight line and collide. Immediately before the collision the particles are moving towards each other. A with speed 3 ms^1 and B with speed 2 ms^1. Immediately after the collision both particles are moving in the same direction, one with speed 3 ms^1 and the other with speed 2 ms^1.
b)Find the ratio m1:m2.
It is given that after the collision both particles are moving in the same direction, one with speed 3 ms^1 and the other with speed 2 ms^1. But how to know the direction is towards the left or to the right? And how to know which has speed of 3 or 2 ms^1?
I couldn't get the correct answer 5:1
Because B has a higher speed after the collision it must have had a lower momentum (as long as the particles are moving towards each other for the collision), so we now know that A will not have changed direction and B will have changed direction.
So from an intial system of (dots are gaps):
A..........B (particles)
>.......< (direction)
3..........2 (speed)
we've now got:
A.........B
>.......>
2.........3
Momentum is always conserved. If A has a mass of a and B has a mass off b the intial momentum must have been 3a2b (B's momentum is negative because B is moving in the opposite direction) and a momentum afterwards of 2a+3b
So:
3a2b=2a+3b
which can be rearranged to give
a=5b
That looks quite difficult because I didn't write it very concisely, but it is a quick and easy way of working it out. If you can't see a shortcut such as this I would do it as Francis suggested, but in many (possibly the majority) cases you can find a shortcut such as this and save some time. 
Civ217
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 08072006 16:57
Firstly thank you everyone for helping.
(Original post by Harr)
Because B has a higher speed after the collision it must have had a lower momentum (as long as the particles are moving towards each other for the collision), so we now know that A will not have changed direction and B will have changed direction.
(Original post by FWoodhouse)
Impulse is change in momentum, so you need to find its momentum after the collision with the floor. You can use v^{2} = u^{2} + 2as to find this; you know that v = 0 at s = 1.2.
(Original post by silent ninja)
400*12 = 400*v1 + 500*v2
also we've been told that, v2v1= 0.5*12. solve.
.
Thank you for everyone again. 
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 08072006 17:20
(Original post by Civ217)
I'm often confused at the equation Impulse = m(vu), because sometimes I get a negative impulse..does that mean that I'm wrong. In this case, impulse = 0.06(5.6(4.849) and I get the correct answer but why is there a negative sign with the value 4.849 and why v is not 4.849 but 5.6..this is always the thing that I am confused.
In general, questions will usually ask for the magnitude of the impulse, so that you don't have to worry about signs. 
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 08072006 17:35
(Original post by Civ217)
Sorry, I know how to do it now but still don't understand it, how do we know that B will have changed its direction?
From the previous section we know that B's speed increases.
If B kept on going in the same direction after the collision its speed would decrease. This is because some of its momentum would have to be transferred to A to cause A to change direction, so it would have a lower momentum and therefore a lower speed. You may want to think of this in terms of kinetic energy instead of momentum, but the effect is the same (if I am just confusing you by bringing kinetic energy in, then just ignore this sentence).
As B's speed has increased it cannot have continued in the same direction after the collision. 
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 08072006 19:10
(Original post by Civ217)
I see..but it is given that the difference is 0.5 u ms^1. From your equation, it seems that u=12, why is it so?
Thank you for everyone again.
The question says
Each time that a moving truck hits a stationary truck at a speed of u ms^1, the difference in the speeds of the two trucks after the collision is 0.5 u ms^1. 
Civ217
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 09072006 16:10
Thank you everyone.
(Original post by FWoodhouse)
Impulse is a vector quantity, and so its sign indicates the direction in which the impulse has acted. In this instance it is positive, because the directionchanging force has acted upwards, which is the direction you have taken to be positive. Likewise, velocity is a vector quantity, and again its sign gives you the direction in which it is acting  you have taken downwards to be negative, so u is negative and v is positive.
In general, questions will usually ask for the magnitude of the impulse, so that you don't have to worry about signs.
A tennis ball of mass 56g, moving horizontally, is struck by a racquet so that its velocity changes from 20 ms^1 east to 30 ms^1 west. i)Find the magnitude of the impulse to the ball by the racquet.
At a later instant the ball is travelling horizontally at 25 ms^1 when it hits the net. This reduces its horizontal velocity to zero in 0.2 s. ii)What constant force does the net exert on the ball?
So, for part i, I thought u should be 20 and v should be 30 so
impulse
= m(vu)
= 0.056(3020) [Or should u be 30 and v be 20, I'm confused at this]
= 2.8 Ns [The answer given is 2.8 Ns, does it matter that I get the negative sign]
For part ii, I thought v is 0 and u is 25 so
F* t = m(vu)
F = 0.056(0(25) / 20
= 7N
In short, I'm confused with what value should we put as u or v and whether to include the negative sign.
Silent ninja, I will do the question later as I have no time to now. And Harr, I understand it now.
Thank you all again. 
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 09072006 17:09
(Original post by Civ217)
A tennis ball of mass 56g, moving horizontally, is struck by a racquet so that its velocity changes from 20 ms^1 east to 30 ms^1 west. i)Find the magnitude of the impulse to the ball by the racquet.
At a later instant the ball is travelling horizontally at 25 ms^1 when it hits the net. This reduces its horizontal velocity to zero in 0.2 s. ii)What constant force does the net exert on the ball?
So, for part i, I thought u should be 20 and v should be 30 so
impulse
= m(vu)
= 0.056(3020) [Or should u be 30 and v be 20, I'm confused at this]
= 2.8 Ns [The answer given is 2.8 Ns, does it matter that I get the negative sign]
(Original post by Civ217)
For part ii, I thought v is 0 and u is 25 so
F* t = m(vu)
F = 0.056(0(25) / 20
= 7N
In short, I'm confused with what value should we put as u or v and whether to include the negative sign.
Silent ninja, I will do the question later as I have no time to now. And Harr, I understand it now.
Thank you all again.
Check a mark scheme or something if you're worried.
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