Animated Solution for Physics - Laws of Motion: Two billiard balls of equal mass 30 g strike a rigid wall with same speed of 108 km/h (as shown) but at different angles. If the balls get reflected with the same speed, then the ratio of the magnitude of impulses imparted to ball $a$ and ball $b$ by the wall along x-direction is

Imagine you are standing in front of a rigid wall, watching two identical billiard balls, each with a mass of $30\text{ g}$, hurtling towards it at a blistering speed of $108\text{ km/h}$.

Ball A is straightforward—it strikes the wall head-on, perfectly perpendicular to the surface. Ball B, however, is a bit more stylish. It comes in at a $45^{\circ }$ angle, strikes the wall, and bounces off at the exact same angle. The problem tells us that both balls reflect with their speeds unchanged. Our mission? To find the ratio of the magnitude of the impulses imparted to Ball A and Ball B by the wall along the x-direction.

Impulse and Momentum

Before we dive into the calculations, let's anchor ourselves to the core physics principle at play here: the Impulse-Momentum Theorem.

Impulse ($I$) is defined as the change in momentum ($\Delta p$) of an object. Mathematically, it is expressed as:

Since the wall is perfectly vertical and we are assuming it's frictionless, it can only exert a normal force perpendicular to its surface. This means the impulse will only act along the horizontal axis (the x-direction). Therefore, we only need to concern ourselves with the x-components of the momentum.

Let's focus on Ball A first. It's moving purely along the x-axis.

Its initial momentum is directed towards the wall (let's call this the positive x-direction):

After the perfectly elastic collision, it bounces back with the same speed but in the opposite direction:

Now, we calculate the change in momentum to find the impulse:

The magnitude of the impulse imparted to Ball A is $2mu$.

Now, let's look at Ball B. It strikes the wall at an angle of $45^{\circ }$. Because momentum is a vector, we must resolve it into its x and y components.

The initial momentum in the x-direction is:

After the collision, the ball reflects at the same $45^{\circ }$ angle with the same speed. The x-component of its final momentum is now pointing away from the wall:

Let's calculate the impulse for Ball B:

Since $\cos 45^{\circ }=\frac{1}{\sqrt{2}}$, we can simplify this to:

The magnitude of the impulse imparted to Ball B is $\sqrt{2}mu$.

We now have the impulses for both balls. The final step is to find their ratio:

Notice how the mass ($m$) and the speed ($u$) beautifully cancel out!

So, the ratio of the impulses is $\sqrt{2}:1$.

Did you notice the trap? The problem provided specific numerical values for the mass ($30\text{ g}$) and the speed ($108\text{ km/h}$). However, because we solved the problem algebraically first, we realized that these values were completely unnecessary. They were distractors designed to consume your time in tedious calculations. Always trust the algebra—it reveals the true elegance of physics!