Simple elastic collision If a particle with mass $m$ collides with a wall at right angles, and the collision is perfectly elastic. The particle hits the wall at $v\ ms^{-1}$. There is no friction or gravity.
So the particle will rebound at $-v\ ms^{-1}$?

What will the change in momentum be?
I did:
$$initial\ momentum = final\ momentum$$
$$mv = m(-v)$$
$$mv = -mv$$
But this doesn't seem right because it's like saying $1=-1$?
 A: The initial and final momentum are not the same because the ball is not an isolated system. The wall exerts a force on it. In principle the ball and the wall (and the planet it's connected to!) form an isolated system with a conserved momentum, but you'd have to take into account how much the wall moves after the collision.
The change of momentum is final momentum - initial momentum, and you have the correct values for the initial and final momentum.
A: In presence of a force the momentum is not conserved, and the wall is a potential repulsive force. Instead, the momentum changes from a positive to a negative value, so the difference is positive.
A: Your equation: $\text {initial momentum = final momentum}$, applies only to the total momentum. It does not apply to individual masses separately. 
Here the initial momentum of the mass $m = \text {initial total momentum} = mv$ (since the wall is not moving) 
The final total momentum is the sum of the momenta of the wall and the momentum of the mass $m$ 
The final total momentum  is thus the initial total momentum = $mv = -mv + x$
Change in the momentum of $m = -mv-mv =-2mv$.
Change in the momentum of the wall $= + 2mv$.
Total change in the momentum of the system $= (-2mv + 2mv) = 0$ (by law of conservation of momentum). You may add the units to the quantities.
Comment: The diagram shows the velocity after collision as $-v \:\mathrm{ms^{-1}}$ with an arrow pointing to the left. That would be incorrect. $-mv$ with arrow pointing to the right or $mv$ with arrow pointing to the left would be correct.
