Problem 59 - College Physics

Question

A mass \(m\) is placed at the rim of a frictionless hemispherical bowl with a radius \(R\) and released from rest (Figure 8.49 ). It then slides down and undergoes a perfectly elastic collision with a second mass \(3 m\) sitting at rest at the bottom of the bowl. (a) What are the direction and speed of each mass just after the collision? (b) In terms of \(R\), to what maximum height will each mass travel after the collision?

Step-by-Step Solution

Answer

After collision, mass m moves backward with velocity \(-\frac{v}{2}\) and mass 3m moves forward with \(\frac{v}{2}\). Each mass reaches a height of \(\frac{R}{4}\).

Step 1: Identify the Initial Conditions

The initial velocity of mass \(m\) is zero since it is released from rest at the rim of the bowl. Mass \(3m\) is also initially at rest. At the highest point, the potential energy for mass \(m\) is \(mgR\) (where \(g\) is the acceleration due to gravity). The total energy will be conserved since the system is frictionless and the collision is perfectly elastic.

Step 2: Calculate the Speed of Mass \(m\) at the Bottom of the Bowl

By conservation of mechanical energy, the potential energy of mass \(m\) at the top is fully converted into kinetic energy at the bottom. \[ mgR = \frac{1}{2}mv^2 \] Solving for \(v\), we find: \[ v = \sqrt{2gR} \].

Step 3: Apply the Conservation of Momentum for the Collision

Before the collision, only mass \(m\) has momentum. The initial momentum is \(mv\). Let the velocity of mass \(m\) after the collision be \(v_1\) and the velocity of mass \(3m\) be \(v_2\). Apply conservation of momentum: \[ mv = mv_1 + 3mv_2 \] Simplifying, we have: \[ v = v_1 + 3v_2 \].

Step 4: Use the Coefficient of Restitution for an Elastic Collision

The coefficient of restitution for an elastic collision is 1. Hence: \[ v_2 - v_1 = v \].

Step 5: Solve the System of Equations for Velocities after Collision

From steps 3 and 4, we have two equations: \[ v = v_1 + 3v_2 \] \[ v_2 - v_1 = v \] Solve these simultaneously. Substituting \(v_2 = v + v_1\) into the momentum equation gives: \[ v = v_1 + 3(v + v_1) \] Simplifying: \[ v = 4v_1 + 3v \] \( v_1 = -\frac{1}{2}v \) and \(v_2 = \frac{1}{2}v\). This means mass \(m\) moves back up at \(\frac{v}{2}\) and mass \(3m\) moves forward at \(\frac{v}{2}\).

Step 6: Calculate the Maximum Heights Reached by Both Masses

For mass \(m\), using conservation of energy again: \[ \frac{1}{2}m(v_1)^2 = mgR_1 \] \[ \frac{1}{2}m\left(\frac{v}{2}\right)^2 = mgR_1 \] Solving: \[ R_1 = \frac{R}{4} \].The same logic applies for mass \(3m\): \[ \frac{1}{2}\cdot 3m(v_2)^2 = 3mgR_2 \] \[ \frac{1}{2}\cdot 3m\left(\frac{v}{2}\right)^2 = 3mgR_2 \] Simplifying: \[ R_2 = \frac{R}{4} \].

Key Concepts

Elastic Collision

In the context of the problem, an elastic collision is key to understanding the interaction between the two masses. An elastic collision is a type of collision where the total kinetic energy and the total momentum of the system are conserved. This means that, unlike inelastic collisions, no kinetic energy is transformed into other forms like heat or sound.

When the small mass, which we'll call mass \(m\), hits the larger stationary mass \(3m\) at the bottom of the bowl, the collision is said to be perfectly elastic. Because of this, both momentum and kinetic energy are preserved.

• All original kinetic energy before the collision is accounted for after the collision.
• The direction of momentum remains consistent with the directions and selected coordinate system.

The conservation of these quantities allows us to set up equations in the problem that help solve for the speeds and directions of both masses after the collision.

Momentum Conservation

Momentum Conservation is a principle that states that if no external forces act on a system, the total momentum of the system remains constant. It's crucial for solving collision problems like the one in the exercise. Momentum is calculated as the product of mass and velocity \((p = mv)\).

Before the collision, the total momentum is the momentum of the moving mass \(m\), which is descending with velocity \(\sqrt{2gR}\). This is given as \(mv\). The stationary mass \(3m\) has no initial momentum since it is at rest.

• Total initial momentum: \(mv\)
• Let post-collision velocities of \(m\) and \(3m\) be \(v_1\) and \(v_2\) respectively.

Using the law of momentum conservation: \[ mv = mv_1 + 3mv_2 \] This equation helps us determine the velocities of the masses after they collide. Without loss of any external force, momentum remains constant throughout the interaction.

Potential Energy

Potential energy is the stored energy of an object due to its position in a force field, most commonly a gravitational field. This energy is defined for mass \(m\) at the top of the bowl since it is positioned at height \(R\). Gravitational potential energy is given by \(mgh\), where \(g\) is the acceleration due to gravity and \(h\) is the height.

In this exercise, the potential energy for the initial state of mass \(m\) is \(mgR\), since the mass is at the rim of the bowl. When it slides to the bottom of the bowl, all this potential energy converts into kinetic energy. This is a demonstration of the conservation of mechanical energy, applying particularly to vertical motion in gravitational fields:

• Initial Potential Energy: \(mgR\)
• Conversion of Potential Energy to Kinetic Energy at bottom: \(\frac{1}{2}mv^2\)

This conversion is what gives the descending mass \(m\) its speed \(\sqrt{2gR}\) when it reaches the bottom and is critical for calculating the motion post-collision.

Kinetic Energy

Kinetic energy is the energy of motion, calculated by \(\frac{1}{2}mv^2\), where \(m\) is mass and \(v\) is velocity. It plays an essential role in the conservation laws that allow us to solve this collision problem. Initially, both masses have no kinetic energy, as they are at rest.

As mass \(m\) descends, potential energy converts into kinetic energy. By the time it reaches the bottom of the bowl, its entire potential energy has morphed into kinetic energy:

• Kinetic Energy at bottom for mass \(m\): \(\frac{1}{2}mv^2\)
• Post-collision System: Both momentum and kinetic energy must be conserved.

After the elastic collision, both masses gain kinetic energy: mass \(m\) moves back with velocity \(\frac{v}{2}\) while \(3m\) moves forward with the same velocity \(\frac{v}{2}\). This division and conservation of kinetic energy ensure that the problem is solved within the constraints of mechanical energy conservation.