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# M1 questions watch

1. I don't understand two parts of the January '09 paper.

The first one:
Two particles A and B are moving on a smooth horizontal plane. The mass of A is km, where 2 < k < 3 , and the mass of B is m. The particles are moving along the same straight line, but in opposite directions, and they collide directly. Immediately before they collide
the speed of A is 2u and the speed of B is 4u. As a result of the collision the speed of A is halved and its direction of motion is reversed.
(a) Find, in terms of k and u, the speed of B immediately after the collision.
(b) State whether the direction of motion of B changes as a result of the collision, explaining your answer.

I worked out part (a) as u(3k - 4) = v using the conservation of momentum, but I don't understand part (b). The mark scheme says that "since k > 2, v > 0, and the direction of motion is reversed", but I can't make any sense of this.

Also, I don't understand parts (b) and (c) of this question:

One end of a light inextensible string is attached to a block P of mass 5 kg. The block P is held at rest on a smooth fixed plane which is inclined to the horizontal at an angle, where sinα = 3/5. The string lies along a line of greatest slope of the plane and passes over a smooth light pulley which is fixed at the top of the plane. The other end of the string is attached to a light scale pan which carries two blocks Q and R, with block Q on top of block R, as shown in Figure 3. The mass of block Q is 5 kg and the mass of block R is 10 kg. The scale pan hangs at rest and the system is released from rest. By modelling the blocks as particles, ignoring air resistance and assuming the motion is uninterrupted,
find
(a) (i) the acceleration of the scale pan,
(ii) the tension in the string,
(b) the magnitude of the force exerted on block Q by block R,
(c) the magnitude of the force exerted on the pulley by the string.

Again, I tried consulting the mark scheme. For part (c) it says:
F=2Tcos((90 - a)/2)
=12gcos26.56
=105N

I understand why we take the angle as 90 - a, but the rest of it confuses me.

Help would be much appreciated.
2. For your first question, you need to examine the total momentum before and after the collision.

-----Before Collision-----

2kmu - 4mu

-----After collision-----

It is given that the speed of A is halved, then, since momentum is conserved we have:

-kmu + (3kmu - 4mu)

-kmu + mu(3k - 4)

So, the speed of B is given by v = u(3k - 4)

Then, if k is 2 < k < 3, the momentum of B after the collision is mu(3k - 4) can be said to be positive since 3(2) > 4. This would mean the direction of B has indeed been reversed.
Remember that momentum is a vector quantity.
3. It would be difficult to help you on the second question without drawing diagrams for you. But if you want to solve it, draw free-body force diagrams for both of the block systems and remember that ma is resultant force of the system on both blocks.
4. Thanks.

I've drawn the diagrams, it's just the "magnitude of the force exerted" parts I didn't understand. Thanks for the help.
5. Ok for the final part, where the string passes over the pulley. We have an angle 90 - a between both of the strings. Imagine if we draw a line through the midpoint of that angle and resolve the tension due to each of the block systems in the direction of that line, angle theta = (90 - a) / 2. Now it can modelled as a single string with a weight 2Tcos((90 - a) / 2).

Hope that helps.

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Updated: January 14, 2010
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