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. 4$\ ms^{-1}$
• B. 8$\ ms^{-1}$
• 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: D) 4 ms^-1