Question

A 250 g ball moving horizontally with a velocity 10.0 m/s strikes at point P of a 720 g frictionless hemisphere, which is at rest on a frictionless horizontal tabletop. Assuming the coefficient of restitution to be 0.8, calculate the velocity of the hemisphere after the collision.

Answer 1

3.27 m/s

Answer 2

To solve this problem, we need to apply the principles of conservation of linear momentum and the definition of the coefficient of restitution.

1. Define Variables and Given Data:

• Mass of the ball, $m_1 = 250 \text{ g} = 0.250 \text{ kg}$
• Initial velocity of the ball, $u_1 = 10.0 \text{ m/s}$ (horizontally)
• Mass of the hemisphere, $m_2 = 720 \text{ g} = 0.720 \text{ kg}$
• Initial velocity of the hemisphere, $u_2 = 0 \text{ m/s}$ (at rest)
• Coefficient of restitution, $e = 0.8$
• Angle of impact, $\theta = 37^\circ$. This is the angle the normal to the surface at point P makes with the horizontal.

Let $v_1$ be the final velocity of the ball and $v_2$ be the final velocity of the hemisphere. Since the hemisphere is on a frictionless horizontal tabletop, it can only move horizontally. Therefore, its final velocity will be purely horizontal, i.e., $\vec{v_2} = v_{2x} \hat{i}$.

2. Apply Conservation of Linear Momentum in the Horizontal (x) Direction: Since there are no external horizontal forces acting on the ball-hemisphere system, the total horizontal momentum is conserved. $m_1 u_{1x} + m_2 u_{2x} = m_1 v_{1x} + m_2 v_{2x}$ Given $u_{1x} = u_1 = 10 \text{ m/s}$ and $u_{2x} = 0$. $m_1 u_1 = m_1 v_{1x} + m_2 v_{2x}$ $0.250 \times 10 = 0.250 v_{1x} + 0.720 v_{2x}$ $2.5 = 0.250 v_{1x} + 0.720 v_{2x}$ (Equation 1)

3. Apply the Coefficient of Restitution along the Normal Direction: The normal direction (line of impact) is perpendicular to the tangent at point P, which means it is along the radius connecting the center of the hemisphere to P. From the diagram, this normal makes an angle $\theta = 37^\circ$ with the horizontal.

The coefficient of restitution $e$ is defined as the ratio of the relative velocity of separation to the relative velocity of approach along the line of impact (normal). $e = \frac{(v_{2n} - v_{1n})}{(u_{1n} - u_{2n})}$

Let's find the components of velocities along the normal direction:

• Initial normal velocity of the ball: $u_{1n} = u_1 \cos \theta = 10 \cos 37^\circ$
• Initial normal velocity of the hemisphere: $u_{2n} = 0$
• Final normal velocity of the ball: $v_{1n} = v_{1x} \cos \theta + v_{1y} \sin \theta$
• Final normal velocity of the hemisphere: $v_{2n} = v_{2x} \cos \theta + v_{2y} \sin \theta$. Since $v_{2y} = 0$, $v_{2n} = v_{2x} \cos \theta$.

Substitute these into the restitution equation: $e (u_1 \cos \theta - 0) = (v_{2x} \cos \theta) - (v_{1x} \cos \theta + v_{1y} \sin \theta)$ $e u_1 \cos \theta = v_{2x} \cos \theta - v_{1x} \cos \theta - v_{1y} \sin \theta$ (Equation 2)

4. Relate Velocities using Impulse: The impulse of collision acts along the normal direction. Let the magnitude of the impulse be $J$. For the ball, the impulse is $-J$ along the normal. For the hemisphere, the impulse is $+J$ along the normal. However, the hemisphere is constrained to move horizontally. The vertical component of the impulse on the hemisphere is absorbed by the tabletop. Only the horizontal component of the impulse from the ball causes horizontal motion of the hemisphere.

• Change in momentum of hemisphere in x-direction: $m_2 v_{2x} - m_2 u_{2x} = J_x = J \cos \theta$ $m_2 v_{2x} = J \cos \theta$ (Equation 3)

• Change in momentum of ball in x-direction: $m_1 v_{1x} - m_1 u_{1x} = -J_x = -J \cos \theta$ $m_1 v_{1x} - m_1 u_1 = -J \cos \theta$ (Equation 4)

• Change in momentum of ball in y-direction: $m_1 v_{1y} - m_1 u_{1y} = -J_y = -J \sin \theta$ $m_1 v_{1y} - 0 = -J \sin \theta$ (Equation 5)

5. Solve the System of Equations: From Equation 3, $J = \frac{m_2 v_{2x}}{\cos \theta}$. Substitute this into Equation 5: $m_1 v_{1y} = - \left( \frac{m_2 v_{2x}}{\cos \theta} \right) \sin \theta = -m_2 v_{2x} \tan \theta$ So, $v_{1y} = -\frac{m_2}{m_1} v_{2x} \tan \theta$ (Equation 6)

From Equation 1, $v_{1x} = \frac{m_1 u_1 - m_2 v_{2x}}{m_1} = u_1 - \frac{m_2}{m_1} v_{2x}$ (Equation 7)

Now substitute $v_{1x}$ (from Eq. 7) and $v_{1y}$ (from Eq. 6) into Equation 2: $e u_1 \cos \theta = v_{2x} \cos \theta - \left( (u_1 - \frac{m_2}{m_1} v_{2x}) \cos \theta + (-\frac{m_2}{m_1} v_{2x} \tan \theta) \sin \theta \right)$ $e u_1 \cos \theta = v_{2x} \cos \theta - u_1 \cos \theta + \frac{m_2}{m_1} v_{2x} \cos \theta + \frac{m_2}{m_1} v_{2x} \frac{\sin^2 \theta}{\cos \theta}$

Multiply the entire equation by $m_1 \cos \theta$ to clear denominators: $m_1 e u_1 \cos^2 \theta = m_1 v_{2x} \cos^2 \theta - m_1 u_1 \cos^2 \theta + m_2 v_{2x} \cos^2 \theta + m_2 v_{2x} \sin^2 \theta$

Rearrange terms to solve for $v_{2x}$: $m_1 e u_1 \cos^2 \theta + m_1 u_1 \cos^2 \theta = v_{2x} (m_1 \cos^2 \theta + m_2 \cos^2 \theta + m_2 \sin^2 \theta)$ $u_1 m_1 \cos^2 \theta (e + 1) = v_{2x} (m_1 \cos^2 \theta + m_2 (\cos^2 \theta + \sin^2 \theta))$ $u_1 m_1 \cos^2 \theta (1 + e) = v_{2x} (m_1 \cos^2 \theta + m_2)$

Finally, solve for $v_{2x}$: $v_{2x} = \frac{u_1 m_1 \cos^2 \theta (1 + e)}{m_1 \cos^2 \theta + m_2}$

6. Substitute Numerical Values: Given: $m_1 = 0.250 \text{ kg}$, $u_1 = 10.0 \text{ m/s}$, $m_2 = 0.720 \text{ kg}$, $e = 0.8$, $\theta = 37^\circ$. We know $\cos 37^\circ \approx 0.8$. So, $\cos^2 37^\circ \approx (0.8)^2 = 0.64$.

$v_{2x} = \frac{10.0 \times 0.250 \times (0.8)^2 \times (1 + 0.8)}{0.250 \times (0.8)^2 + 0.720}$ $v_{2x} = \frac{10.0 \times 0.250 \times 0.64 \times 1.8}{0.250 \times 0.64 + 0.720}$ $v_{2x} = \frac{2.5 \times 0.64 \times 1.8}{0.16 + 0.720}$ $v_{2x} = \frac{1.6 \times 1.8}{0.88}$ $v_{2x} = \frac{2.88}{0.88}$ $v_{2x} \approx 3.2727 \text{ m/s}$

The velocity of the hemisphere after the collision is approximately $3.27 \text{ m/s}$.

The final answer is $\boxed{\text{3.27 m/s}}$

Explanation of the solution:

1. Identify forces and constraints: The collision involves an external force from the table on the hemisphere. This force prevents vertical motion of the hemisphere. Thus, momentum is conserved only in the horizontal direction for the ball-hemisphere system.
2. Define line of impact: The collision force acts along the normal to the surface at point P. This normal makes an angle of $37^\circ$ with the horizontal.
3. Apply Conservation of Horizontal Momentum: $m_1 u_1 = m_1 v_{1x} + m_2 v_{2x}$. This gives one equation relating the final horizontal velocities.
4. Apply Coefficient of Restitution: The coefficient of restitution ($e$) relates the relative velocities along the normal direction before and after the collision: $e = \frac{v_{2n} - v_{1n}}{u_{1n} - u_{2n}}$.
5. Relate impulse and momentum change: The impulse from the collision acts along the normal. For the hemisphere, only the horizontal component of this impulse ($J \cos \theta$) contributes to its horizontal motion, as the vertical component is absorbed by the table. For the ball, the impulse changes both its horizontal and vertical momentum components.
6. Formulate equations: Express the final velocities ($v_{1x}, v_{1y}, v_{2x}$) in terms of the impulse $J$ and initial conditions.
7. Solve the system: Substitute the impulse-derived velocity components into the restitution equation and the momentum conservation equation to solve for the unknown velocity of the hemisphere ($v_{2x}$).
8. Calculate: Plug in the given numerical values to get the final answer.

Answer:

The velocity of the hemisphere after the collision is approximately 3.27 m/s.