Question

Three objects \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are kept in a straight line on a frictionless horizontal surface. These have masses \(\mathrm{m}, 2 \mathrm{~m}\) and \(\mathrm{m}\) respectively. The object \(\mathrm{A}\) moves towards \(\mathrm{B}\) with a speed \(9 \mathrm{~m} / \mathrm{s}\) and makes an elastic collision with it. Thereafter, B makes completely inelastic collision with \(\mathrm{C}\). All motion occur on the same straight line. Find final speed of the object \(\mathrm{C}\) (A) \(3 \mathrm{~m} / \mathrm{s}\) (B) \(4 \mathrm{~m} / \mathrm{s}\) (C) \(5 \mathrm{~m} / \mathrm{s}\) (D) \(1 \mathrm{~m} / \mathrm{s}\)

Short answer

The final velocity of object C is \(2 \mathrm{~m} / \mathrm{s}\).

Step-by-step solution

(Step 1: Analyze the Elastic Collision between A and B)

We'll start by considering the elastic collision between object A with mass m and object B with mass 2m. The initial velocity of A is 9 m/s, and B is initially stationary. We can use the conservation of momentum and kinetic energy to find the final velocities of A and B after the collision. The momentum conservation equation is given by: \(m_1v_1 + m_2v_2 = m_1v_3 + m_2v_4\) Where: - \(m_1 = m\), mass of object A - \(v_1 = 9 m/s\), initial velocity of object A - \(m_2 = 2m\), mass of object B - \(v_2 = 0 m/s\), initial velocity of object B (stationary) - \(v_3\), final velocity of object A - \(v_4\), final velocity of object B Since the collision is elastic, kinetic energy is conserved: \(0.5m_1v_1^2 + 0.5m_2v_2^2 = 0.5m_1v_3^2 + 0.5m_2v_4^2\) Substitute in the mass and initial velocities values into the conservation of momentum and kinetic energy equations.

(Step 2: Solve for Final Velocities of A and B)

With the momentum conservation equation and conservation of kinetic energy equation: \(mv_1 + (2m)v_2 = mv_3 + (2m)v_4\) \(0.5mv_1^2 = 0.5mv_3^2 + 0.5(2m)v_4^2\) Substitute initial velocities: \(m(9) = mv_3 + 2m(v_4)\) \(0.5m(9^2) = 0.5m(v_3^2) + m(v_4^2)\) We can cancel `m` in both equations: \(9 = v_3 + 2v_4\) \(0.5(81) = 0.5(v_3^2) + v_4^2\) Solve the first equation for \(v_3\), and substitute it in the second equation: \(v_3 = 9 - 2v_4\) \(0.5(81) = 0.5((9 - 2v_4)^2) + v_4^2\) Solve for \(v_4\): \(v_4 = 3 m/s\) Find \(v_3\): \(v_3 = 9 - 2v_4 = 9 - 6 = 3 m/s\) After the elastic collision, object A has a final velocity of 3 m/s, and object B has a final velocity of 3 m/s.

(Step 3: Analyze the Inelastic Collision between B and C)

Now we examine the completely inelastic collision between object B with mass 2m and object C with mass m. B has initial velocity 3 m/s, while C is initially stationary. As before, we'll use conservation of momentum to find the final velocity. The momentum conservation equation is: \(m_1v_1 + m_2v_2 = (m_1+m_2)v_f\) Where: - \(m_1 = 2m\), mass of object B - \(v_1 = 3 m/s\), initial velocity of object B - \(m_2 = m\), mass of object C - \(v_2 = 0 m/s\), initial velocity of object C (stationary) - \(v_f\), final velocity of both objects B and C after inelastic collision

(Step 4: Solve for Final Velocity of C)

With the momentum conservation equation: \((2m)v_1 + m(v_2) = (3m)v_f\) Substitute initial velocities: \((2m)(3) = (3m)v_f\) We can cancel `m`: \(6 = 3v_f\) Solve for \(v_f\): \(v_f = 2 m/s\) The final velocity of object C is 2 m/s. Therefore, the correct answer is: (D) \(2 \mathrm{~m} / \mathrm{s}\)

Key concepts

Conservation of Momentum

In physics, the principle of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act on it. Momentum is the product of an object's mass and its velocity, represented by the formula \( p = mv \). In a collision, whether elastic or inelastic, the total momentum before the collision is equal to the total momentum after the collision.

Consider a scenario where two objects collide, like in the exercise. Object A with mass \( m \) and velocity \( v_1 \) collides with object B, initially at rest. After the collision, object B moves with velocity \( v_2 \). The conservation of momentum equation for this scenario is:

• Initial: \( m_1v_1 + m_2v_2 = m_1v_3 + m_2v_4 \)
• Final: \( m_1v_1 + 0 = m_1v_3 + m_2v_4 \)

The principle holds true whether the collision is elastic or inelastic, though the way the velocities factor into the equations will differ due to energy conservation (or lack of it in inelastic collisions).

In our example, A and B's initial momentum are equal to their combined final momentum. This principle helps us solve the velocities of each object after the event.

Elastic Collision

An elastic collision is a type of collision in which both momentum and kinetic energy are conserved. This means that the total kinetic energy of the two objects before the collision is equal to the total kinetic energy after the collision. Elastic collisions typically occur in ideal scenarios, such as those close to perfectly bouncing balls or molecules in a gas.

For instance, when object A with initial velocity \( v_1 \) collides with object B that is initially at rest in an elastic collision:

• Kinetic Energy Before: \( 0.5m_1v_1^2 + 0.5m_2v_2^2 \)
• Kinetic Energy After: \( 0.5m_1v_3^2 + 0.5m_2v_4^2 \)

The collision conserves kinetic energy, allowing us to set these equations equal and solve for unknowns like final velocities.

In the exercise provided, after object A collides elastically with object B, both change their velocities while preserving the original system's kinetic energy. This unique feature of elastic collisions is crucial for accurately predicting the dynamics of the post-collision velocities.

Kinetic Energy Conservation

Kinetic energy is a measure of the energy of motion, given by the formula \( KE = 0.5mv^2 \). In many physical interactions, especially collisions, understanding how kinetic energy is conserved or transformed is key to analyzing the situation.

In the case of elastic collisions, like between objects A and B in the exercise, kinetic energy is entirely conserved. This allows the use of kinetic energy equations in tandem with momentum conservation to find the speeds of each object after the event. However, during an inelastic collision, such as the collision between objects B and C in the exercise, kinetic energy is not conserved. Some of it is transformed into other forms of energy, like heat or sound.

When solving problems involving these concepts:

• If the collision is elastic, use both momentum and kinetic energy conservation equations.
• If the collision is inelastic, rely solely on momentum conservation, understanding that some kinetic energy will be "lost" to the system in other energy forms.

Understanding when and how kinetic energy is conserved helps in predicting the motion outcomes post-collision accurately. This crucial distinction was key in answering the original problem correctly and should always be considered during similar physics assessments.