Question

A body of mass 1 kg is suspended by a weightless string which passes over a frictionless pulley of mass 2 kg as shown in the figure. The mass is released from a height of 1.6m from the ground. With what velocity does it strike the ground ?

• A. 8$\ ms^{-1}$
• B. 4$\ ms^{-1}$
• C. $4\sqrt 2\ ms^{-1}$
• D. 16$\ ms^{-1}$

Answer 1

The Correct Option is B

Let \(m\) be the mass of the block (1 kg), \(M\) be the mass of the pulley (2 kg), and \(h\) be the height (1.6 m).

The potential energy lost by the block is converted into kinetic energy of the block and rotational kinetic energy of the pulley.

Loss in Potential Energy (PE) = \(mgh = 1 \times 9.8 \times 1.6\) J

Kinetic Energy (KE) of the block = \(\frac{1}{2}mv^2 = \frac{1}{2} \times 1 \times v^2\)

Moment of Inertia (I) of the pulley (assuming it's a disc) = \(\frac{1}{2}MR^2\)

Rotational Kinetic Energy (RKE) of the pulley = \(\frac{1}{2}I\omega^2 = \frac{1}{2} (\frac{1}{2}MR^2) (\frac{v}{R})^2 = \frac{1}{4}Mv^2 = \frac{1}{4} \times 2 \times v^2 = \frac{1}{2}v^2\)

By conservation of energy:

Loss in PE = KE of the block + RKE of the pulley

\(mgh = \frac{1}{2}mv^2 + \frac{1}{4}Mv^2\)

\(1 \times 9.8 \times 1.6 = \frac{1}{2} \times 1 \times v^2 + \frac{1}{4} \times 2 \times v^2\)

\(15.68 = \frac{1}{2}v^2 + \frac{1}{2}v^2\)

\(15.68 = v^2\)

\(v = \sqrt{15.68} \approx 3.96\) m/s. Approximating, we get v ≈ 4 m/s.

Since \(g=10\) is used:

\(1 \times 10 \times 1.6 = \frac{1}{2} \times 1 \times v^2 + \frac{1}{4} \times 2 \times v^2\)

\(16 = \frac{1}{2}v^2 + \frac{1}{2}v^2\)

\(16 = v^2\)

\(v = \sqrt{16} = 4\) m/s.

Answer: B) 4 ms^-1

Answer 2

This is a problem of motion with a pulley system. We can solve this problem using energy conservation principles, as the system is frictionless.
Given:
Mass of the hanging body, \( m_1 = 1 \, \text{kg} \)
Mass of the pulley, \( m_2 = 2 \, \text{kg} \)
Height from which the body is released, \( h = 1.6 \, \text{m} \)
The system is frictionless, and the pulley is massless.
When the body is released, it accelerates downward, and the pulley rotates. As there is no friction, the total mechanical energy of the system is conserved. We can equate the potential energy lost by the hanging mass to the kinetic energy gained by the mass and the pulley. Step 1: Mechanical Energy Conservation The initial potential energy of the hanging body is converted into the kinetic energy of the body and the kinetic energy of the rotating pulley. - Initial potential energy of the body: \[ PE = m_1 g h = 1 \times 9.8 \times 1.6 = 15.68 \, \text{J} \] - The kinetic energy of the body when it strikes the ground: \[ KE_{\text{body}} = \frac{1}{2} m_1 v^2 \] where \( v \) is the velocity of the body when it strikes the ground. - The kinetic energy of the rotating pulley: \[ KE_{\text{pulley}} = \frac{1}{2} I \omega^2 \] where \( I = \frac{1}{2} m_2 r^2 \) is the moment of inertia of the pulley, and \( \omega \) is the angular velocity of the pulley. Since the string is inextensible, the linear velocity of the body and the tangential velocity of the pulley's rim are equal: \[ v = r\omega \quad \Rightarrow \quad \omega = \frac{v}{r} \] Thus, the kinetic energy of the pulley becomes: \[ KE_{\text{pulley}} = \frac{1}{2} \times \frac{1}{2} m_2 r^2 \times \left( \frac{v}{r} \right)^2 = \frac{1}{4} m_2 v^2 \]
Step 2: Total Energy Conservation At the starting point, all the energy is potential energy, and at the final point, it is the sum of the kinetic energy of the body and the pulley: \[ PE_{\text{initial}} = KE_{\text{body}} + KE_{\text{pulley}} \] \[ 15.68 = \frac{1}{2} m_1 v^2 + \frac{1}{4} m_2 v^2 \] Substitute the values: \[ 15.68 = \frac{1}{2} \times 1 \times v^2 + \frac{1}{4} \times 2 \times v^2 \] \[ 15.68 = \frac{1}{2} v^2 + \frac{1}{2} v^2 \] \[ 15.68 = v^2 \] \[ v = \sqrt{15.68} \approx 4 \, \text{m/s} \] Thus, the velocity with which the body strikes the ground is approximately 4 m/s.