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# Momentum watch

1. 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.
2. (Original post by Civ-217)
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
There might be a neater way, but the easy way is just to try all 4 possible situations, writing down conservation of momentum for each one (be careful with your signs - decide which way positive is and which way negative is). You should find that for two of them you end up with one of the particles having zero mass. A third one will give you a negative mass for one of the particles. Therefore, there's only one possible situation that works.

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.
Impulse is change in momentum, so you need to find its momentum after the collision with the floor. You can use v2 = u2 + 2as to find this; you know that v = 0 at s = 1.2.
3. (Original post by Civ-217)
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.
Is restitution in your syllabus? This is basically it.
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, v2-v1= 0.5*12. solve.
do the same for the second truck colliding with third truck.

500*v2= 500*v3 + 300*v4
v4-v3= 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.
4. (Original post by Civ-217)
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
In any collision both particles will experience a change in velocity. Therefore if A is initially moving at 3m/s it will not be moving at 3m/s after the collision. So after the collision A will have a speed of 2m/s and B a speed of 3m/s.

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 3a-2b (B's momentum is negative because B is moving in the opposite direction) and a momentum afterwards of 2a+3b

So:
3a-2b=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.
5. 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.
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?

(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 v2 = u2 + 2as to find this; you know that v = 0 at s = 1.2.
I'm often confused at the equation Impulse = m(v-u), 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.

(Original post by silent ninja)
400*12 = 400*v1 + 500*v2
also we've been told that, v2-v1= 0.5*12. solve.
.
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.
6. (Original post by Civ-217)
I'm often confused at the equation Impulse = m(v-u), 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.
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 direction-changing 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.
7. (Original post by Civ-217)
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?
I probably wasn't very clear. I'll try and explain again. If it's still not clear then just tell me and I'll try and think of a different way of explaining it.

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.
8. (Original post by Civ-217)
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.
so u is the speed at which a truck hits a stationary truck. for the first when it hits the second truck it approaches at u=12, for the second truck when it hits the third, it approaches at u= v2 as above.
9. 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 direction-changing 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.
I will tell you why I get confused with this by the following question.

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(v-u)
= 0.056(-30-20) [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(v-u)
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.
10. (Original post by Civ-217)
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(v-u)
= 0.056(-30-20) [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]
Blah, impulse is usually given as positive. The racquet is moving, relative to the player, 'forwards', and it's the racquet that strikes the ball. If it then asked what the impulse of the ball on the racquet was, you would give -2.8Ns. But to be honest, this is all nitpicking, and they will ask for the magnitude of the impulse anyway - and magnitude is always positive (or zero).

(Original post by Civ-217)
For part ii, I thought v is 0 and u is -25 so
F* t = m(v-u)
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.
Obviously, u is initial speed and v is final speed. I don't understand why you've made u negative, but this is fine too. Direction only matters when you have things moving in different directions. Here it really doesn't matter whether you take u to be positive or negative, as long as you're consistent. If you're really worried and you get a negative answer for a force, just write something like "Force is 7N against motion of the ball" afterwards - making it clear that all you've done is changed the direction by putting that - in front of it.

Check a mark scheme or something if you're worried.

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