A doubt about collisions and conservation of momentum

Question

Consider two objects $B_1$ and $B_2$ of equal mass $m$. The object $B_2$ is stationary, while $B_1$ is moving towards $B_2$ with velocity $\mathbf{v}$. Suppose that after the collision, $B_1$ and $B_2$ have velocities $\mathbf{w}_1$ and $\mathbf{w}_2$ that are collinear with $\mathbf{v}$ and, moreover, point in the same direction as $\mathbf{v}$. Show that $|\mathbf{w}_1|\leq|\mathbf{v}|/2$. (This is a very simple physics problem, not a mathematics problem!)

Here is what I got: By conservation of momentum, we have: $$ mv = mw_1 + mw_2 \\ v = w_1 + w_2 $$

Since they are collinear and in the same direction, we have $v\ge 0$, $w_1\ge 0$, and $w_2\ge 0$. Now, I am confused how I am supposed to prove that.

It is my understanding that, for every possible value of $w_2$, I'll have a different collision. And so, if we consider the special case where $v=6$ and $w_2 = 1$, then by momentum conservation $w_1 = 5$, we don't have $w_1\le v/2$.

Am I doing something wrong? How do I prove what my book wants?

Answer 1

Here's another way by using symmetry argument:

Switch to a reference frame in which both bodies are moving towards each other with equal speed, i.e. body $B_1$ has velocity $\mathbf{v}/2$ and body $B_2$ has velocity $-\mathbf{v}/2$. Masses of the two bodies are equal. So in this reference frame the collision process is completely symmetric; this symmetry must be preserved subsequent to collision as well, because there is nothing to break the symmetry. In particular (in this frame) subsequent to collision both bodies move away from each with the same speed $w^*$. Note that subsequent to collision direction of motion of each body is reversed.

By energy conservation total kinetic energy of the two bodies cannot be greater than the initial value, i.e. $w^{*2}\leq (v/2)^2$ or $w^*\leq v/2$ (we must have $w^*\geq 0$ because it is a scalar speed). Therefore $\mathbf{w}^*_1=-\alpha \mathbf{v}/2,\mathbf{w}^*_2=\alpha \mathbf{v}/2$, in which $\alpha\in [0,1]$ and $\alpha=1$ corresponds to elastic collision.

Now we switch back to the reference frame in which $B_2$ was stationary prior to collision. This is done by adding velocity $\mathbf{v}/2$ to everything. In this frame then: $\mathbf{w}_1=(1-\alpha) \mathbf{v}/2,\mathbf{w}_2=(1+\alpha) \mathbf{v}/2,\alpha\in [0,1]$. We are automatically guaranteed that both bodies move in the same direction as $\mathbf{v}$. Finally it follows that $w_1=(1-\alpha)(v/2)$ or $w_1\leq v/2$.

Answer 2

The key point is that for w1 > v/2 the projectile should go through the target. This is unphysical.

The same rationale is used to conclude that, in a totally elastic collision with all other parameters as in your exercise, the two masses simply swap their velocities. Note that from mathematicaly work out momentum and kinetic energy conservation the solutions are w1 = v , w2 = 0 as well as w1 = 0 , w2 = v. The first one is discarded for the reason above.

In brief, the condition v1 < or = v2 must hold.

Answer 3

Using equations of conservation of linear momentum and the definition of coefficient of restitution,

For ball B1 we will have maximum (final) speed in case of inelastic collision; minimum speed in elastic collision. For all other values of 'e' the speed will lie in between the elastic and inelastic cases.

Maximum speed at $e=0$ can be calculated as $v/2$

Minimum speed at $e=1$ can be calculated as $0$

Thus we have $|\mathbf{w}_1|\leq|\mathbf{v}|/2$.