Solution to Question 16
1. Analysis of Initial Motion
Both balls are released from a height $h$. Since they fall freely under gravity, they acquire the same speed $u$ just before reaching the ground level.
$$ u = \sqrt{2gh} $$
The heavy steel ball hits the ground first and rebounds elastically. Since it is very heavy compared to the lighter ball, its velocity simply reverses direction without loss of magnitude (wall approximation).
Just before the collision between the balls:
- Heavy Ball (H): Moving upwards with velocity $\vec{v}_H = u \hat{j}$.
- Light Ball (L): Moving downwards with velocity $\vec{v}_L = -u \hat{j}$.
2. Collision Geometry and Dynamics
The line joining the centers (Line of Impact, $\hat{n}$) makes an angle $\theta = 30^\circ$ with the vertical.
We define the unit vectors for the normal ($\hat{n}$) and tangential ($\hat{t}$) directions:
$$ \hat{n} = \sin\theta \hat{i} + \cos\theta \hat{j} $$
$$ \hat{t} = \cos\theta \hat{i} – \sin\theta \hat{j} $$
Velocity Components along the Line of Impact ($\hat{n}$):
The velocity of the heavy ball along the normal is:
$$ v_{H,n} = \vec{v}_H \cdot \hat{n} = (u \hat{j}) \cdot (\sin\theta \hat{i} + \cos\theta \hat{j}) = u \cos\theta $$
The velocity of the light ball along the normal is:
$$ v_{L,n} = \vec{v}_L \cdot \hat{n} = (-u \hat{j}) \cdot (\sin\theta \hat{i} + \cos\theta \hat{j}) = -u \cos\theta $$
Velocity of Separation:
Since the collision is elastic ($e=1$) and the heavy ball acts as an infinite mass, the relative speed along the normal is preserved.
$$ v_{sep} = e \cdot v_{app} $$
$$ v_{L,n}’ – v_{H,n}’ = 1 \cdot (v_{H,n} – v_{L,n}) $$
Since the heavy ball is massive, its velocity remains unchanged: $v_{H,n}’ \approx v_{H,n} = u \cos\theta$.
$$ v_{L,n}’ – u \cos\theta = u \cos\theta – (-u \cos\theta) $$
$$ v_{L,n}’ = 3u \cos\theta $$
Tangential Component:
The impulse acts only along the normal. Therefore, the tangential velocity of the light ball remains unchanged.
$$ v_{L,t} = \vec{v}_L \cdot \hat{t} = (-u \hat{j}) \cdot (\cos\theta \hat{i} – \sin\theta \hat{j}) = u \sin\theta $$
$$ v_{L,t}’ = v_{L,t} = u \sin\theta $$
3. Final Velocity and Range Calculation
The total velocity of the light ball after collision is:
$$ \vec{v}’ = v_{L,n}’ \hat{n} + v_{L,t}’ \hat{t} $$
$$ \vec{v}’ = (3u \cos\theta)(\sin\theta \hat{i} + \cos\theta \hat{j}) + (u \sin\theta)(\cos\theta \hat{i} – \sin\theta \hat{j}) $$
Resolving into $x$ (horizontal) and $y$ (vertical) components:
Horizontal Velocity ($v_x$):
$$ v_x = 3u \cos\theta \sin\theta + u \sin\theta \cos\theta = 4u \sin\theta \cos\theta = 2u \sin(2\theta) $$
Vertical Velocity ($v_y$):
$$ v_y = 3u \cos^2\theta – u \sin^2\theta = u(3\cos^2\theta – \sin^2\theta) $$
Using the identity $\cos 2\theta = \cos^2\theta – \sin^2\theta$:
$$ v_y = u(2\cos^2\theta + (\cos^2\theta – \sin^2\theta)) = u(2\cos^2\theta + \cos 2\theta) $$
Alternatively, simplifying $3\cos^2\theta – \sin^2\theta$:
$$ v_y = u(3(\frac{3}{4}) – \frac{1}{4}) = u(\frac{8}{4}) = 2u \quad (\text{for } \theta=30^\circ) $$
Horizontal Range ($R$):
The collision occurs effectively at the ground level ($y \approx 0$). The range of the projectile is given by:
$$ R = \frac{2 v_x v_y}{g} $$
Substituting the values for $\theta = 30^\circ$:
$$ v_x = 2u \sin(60^\circ) = 2u \frac{\sqrt{3}}{2} = u\sqrt{3} $$
$$ v_y = 2u $$
$$ R = \frac{2 (u\sqrt{3}) (2u)}{g} = \frac{4\sqrt{3} u^2}{g} $$
Since $u = \sqrt{2gh}$, we have $u^2 = 2gh$.
$$ R = \frac{4\sqrt{3} (2gh)}{g} = 8\sqrt{3} h $$
Calculation:
Given $h = 5\sqrt{3}$ m.
$$ R = 8\sqrt{3} \times (5\sqrt{3}) $$
$$ R = 40 \times 3 = 120 \text{ m} $$
Answer: The horizontal range of the lighter ball is 120 m.
1. Analysis of Initial Motion
Both balls are released from a height $h$. Since they fall freely under gravity, they acquire the same speed $u$ just before reaching the ground level. $$ u = \sqrt{2gh} $$
The heavy steel ball hits the ground first and rebounds elastically. Since it is very heavy compared to the lighter ball, its velocity simply reverses direction without loss of magnitude (wall approximation).
Just before the collision between the balls:
- Heavy Ball (H): Moving upwards with velocity $\vec{v}_H = u \hat{j}$.
- Light Ball (L): Moving downwards with velocity $\vec{v}_L = -u \hat{j}$.
2. Collision Geometry and Dynamics
The line joining the centers (Line of Impact, $\hat{n}$) makes an angle $\theta = 30^\circ$ with the vertical. We define the unit vectors for the normal ($\hat{n}$) and tangential ($\hat{t}$) directions: $$ \hat{n} = \sin\theta \hat{i} + \cos\theta \hat{j} $$ $$ \hat{t} = \cos\theta \hat{i} – \sin\theta \hat{j} $$
Velocity Components along the Line of Impact ($\hat{n}$):
The velocity of the heavy ball along the normal is:
$$ v_{H,n} = \vec{v}_H \cdot \hat{n} = (u \hat{j}) \cdot (\sin\theta \hat{i} + \cos\theta \hat{j}) = u \cos\theta $$
The velocity of the light ball along the normal is:
$$ v_{L,n} = \vec{v}_L \cdot \hat{n} = (-u \hat{j}) \cdot (\sin\theta \hat{i} + \cos\theta \hat{j}) = -u \cos\theta $$
Velocity of Separation:
Since the collision is elastic ($e=1$) and the heavy ball acts as an infinite mass, the relative speed along the normal is preserved.
$$ v_{sep} = e \cdot v_{app} $$
$$ v_{L,n}’ – v_{H,n}’ = 1 \cdot (v_{H,n} – v_{L,n}) $$
Since the heavy ball is massive, its velocity remains unchanged: $v_{H,n}’ \approx v_{H,n} = u \cos\theta$.
$$ v_{L,n}’ – u \cos\theta = u \cos\theta – (-u \cos\theta) $$
$$ v_{L,n}’ = 3u \cos\theta $$
Tangential Component:
The impulse acts only along the normal. Therefore, the tangential velocity of the light ball remains unchanged.
$$ v_{L,t} = \vec{v}_L \cdot \hat{t} = (-u \hat{j}) \cdot (\cos\theta \hat{i} – \sin\theta \hat{j}) = u \sin\theta $$
$$ v_{L,t}’ = v_{L,t} = u \sin\theta $$
3. Final Velocity and Range Calculation
The total velocity of the light ball after collision is: $$ \vec{v}’ = v_{L,n}’ \hat{n} + v_{L,t}’ \hat{t} $$ $$ \vec{v}’ = (3u \cos\theta)(\sin\theta \hat{i} + \cos\theta \hat{j}) + (u \sin\theta)(\cos\theta \hat{i} – \sin\theta \hat{j}) $$
Resolving into $x$ (horizontal) and $y$ (vertical) components:
Horizontal Velocity ($v_x$):
$$ v_x = 3u \cos\theta \sin\theta + u \sin\theta \cos\theta = 4u \sin\theta \cos\theta = 2u \sin(2\theta) $$
Vertical Velocity ($v_y$):
$$ v_y = 3u \cos^2\theta – u \sin^2\theta = u(3\cos^2\theta – \sin^2\theta) $$
Using the identity $\cos 2\theta = \cos^2\theta – \sin^2\theta$:
$$ v_y = u(2\cos^2\theta + (\cos^2\theta – \sin^2\theta)) = u(2\cos^2\theta + \cos 2\theta) $$
Alternatively, simplifying $3\cos^2\theta – \sin^2\theta$:
$$ v_y = u(3(\frac{3}{4}) – \frac{1}{4}) = u(\frac{8}{4}) = 2u \quad (\text{for } \theta=30^\circ) $$
Horizontal Range ($R$): The collision occurs effectively at the ground level ($y \approx 0$). The range of the projectile is given by: $$ R = \frac{2 v_x v_y}{g} $$ Substituting the values for $\theta = 30^\circ$: $$ v_x = 2u \sin(60^\circ) = 2u \frac{\sqrt{3}}{2} = u\sqrt{3} $$ $$ v_y = 2u $$ $$ R = \frac{2 (u\sqrt{3}) (2u)}{g} = \frac{4\sqrt{3} u^2}{g} $$ Since $u = \sqrt{2gh}$, we have $u^2 = 2gh$. $$ R = \frac{4\sqrt{3} (2gh)}{g} = 8\sqrt{3} h $$
Calculation: Given $h = 5\sqrt{3}$ m. $$ R = 8\sqrt{3} \times (5\sqrt{3}) $$ $$ R = 40 \times 3 = 120 \text{ m} $$