The first thing to note is that there are no external impulses acting on our system of two particles. This means that
momentum must be conserved, i.e. the momentum of the particles before the collision is equal to the moment of the
particles after the collision. Remember that the equation for momentum is:
$$momentum = mass \times velocity$$

It is always important in questions like this that you have a clear sign convention. In this case, lets take the direction of ??P?? prior
to the collision to be our positive direction and the direction of ??Q?? prior to the collision to be our negative direction.

Lets call the momentum of ??P?? before the collision ??p^{b}_{P}?? (the ??b?? stands for before) and we can write this as the product of the mass of ??P?? and the velocity of ??P??:
$$
p^{b}_{P} = m_{P} \times v^{b}_{P} = 4m \times 2u = 8mu
$$

The momentum of ??P?? after the collision is ??p^{a}_{P}?? (the ??a?? stands for after) and can be written as:
$$
p^{a}_{P} = m_{P} \times v^{a}_{P} = 4m \times -\frac{u}{2} = -2mu
$$

The momentum of ??Q?? before the collision is ??p^{b}_{Q}?? and can be written as:
$$
p^{b}_{Q} = m_{Q} \times v^{b}_{Q} = m \times -5u = -5mu
$$

Finally the moment of ??Q?? after the collision is ??p^{a}_{Q}?? and can be written as:
$$
p^{a}_{Q} = m_{Q} \times v^{a}_{Q} = m \times v_{a}^{Q} = mv_{a}^{Q}
$$

Now that we have calculated the momentum of each particle before and after the collision we an simply set the
momentum before (??p^{b}??) the collision to be equal to the momentum afterwards (??p^{a}??):
$$
p^{b} = p ^{a}
\\
p^{b}_{P} + p^{b}_{Q} = p^{a}_{P} + p^{a}_{Q}
\\
8mu + (-5mu) = -2mu + mv_{a}^{Q}
\\
3mu + 2mu = mv_{a}^{Q}
\\
5mu = mv_{a}^{Q}
\\
\therefore v_{a}^{Q} = 5u
$$

Make sure that in the exam you have clearly stated your sign convention and it is always useful to just clarify
at the end as well, so you could write that the speed of ??Q?? is ??5u?? in the original direction of ??P??.

Question 1. (b)

Here, we have to find the magnitude of the impulse exerted on ??P?? by ??Q?? in the collision. Remember that:
$$impulse = change \, in \, momentum$$

We can simply calculate the impulse experienced by ??P?? by calculating its change in momentum:
$$
I_{P} = p^{a}_{P} - p^{b}_{P}
\\
I_{P} = -2mu - 8mu
\\
\therefore I_{P} = 10mu
$$

Note that the question only asks for the magnitude, so if we had calculated the impulse to be ??-10mu?? we would
have just put ??10mu?? for the final answer, in this case it does not matter as the magnitude of ??10mu?? is ??10mu?? as well.

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