Question

Bob \(B\) of mass \(m\) at rest is hanging vertically from the ceiling by a massless string of length \(10\,\text{m}\), as shown in the figure. Point mass \(A\) of mass \(m\) travelling horizontally with speed \(10\,\text{m s}^{-1}\) collides with the bob \(B\) elastically. The bob \(B\) rises to a height \(h\) after the collision. Taking acceleration due to gravity \(g=10\,\text{m s}^{-2}\) and neglecting the size of the bob, the value of \(h\) is:

• A. \(2.5\,\text{m}\)
• B. \(8\,\text{m}\)
• C. \(7\,\text{m}\)
• D. \(5\,\text{m}\)

Hint:

For a perfectly elastic collision between two identical masses, the velocities are exchanged. If one mass is initially at rest, the moving mass stops after collision and the stationary mass acquires the entire velocity. After collision, use \[ \frac12 mv^2=mgh \] to determine the maximum height reached.

Answer 1

The Correct Option is D

Concept:

• In a one-dimensional perfectly elastic collision between two identical masses, the velocities are exchanged.

• Linear momentum is conserved.

• Kinetic energy is also conserved.

• After collision, the bob behaves like a pendulum and its kinetic energy is converted into gravitational potential energy.

Step 1: Write the initial conditions of the collision
Mass of particle \(A\) \[ m \] Mass of bob \(B\) \[ m \] Initial velocity of \(A\) \[ u_A=10\,\text{m s}^{-1} \] Initial velocity of \(B\) \[ u_B=0 \] The collision is perfectly elastic.

Step 2: Apply the result for elastic collision of equal masses
For a head-on elastic collision between two equal masses, \[ v_A=u_B \] and \[ v_B=u_A \] Therefore, \[ v_A=0 \] and \[ v_B=10\,\text{m s}^{-1} \] Thus immediately after collision the bob moves with speed \[ \boxed{10\,\text{m s}^{-1}} \]

Step 3: Determine the kinetic energy of the bob after collision
The kinetic energy possessed by the bob is \[ K=\frac12 mv_B^2 \] Substituting \(v_B=10\,\text{m s}^{-1}\), \[ K = \frac12 m(10)^2 \] \[ K = 50m \] \[ K=50m\ \text{J} \]

Step 4: Use conservation of mechanical energy during upward motion
As the bob rises upward, its kinetic energy is converted completely into gravitational potential energy. At the highest point, \[ \text{K.E.}=0 \] and \[ \text{P.E.}=mgh \] Using conservation of energy, \[ \frac12 mv^2=mgh \] Substituting values, \[ \frac12 m(10)^2 = m(10)h \] \[ 50m = 10mh \]

Step 5: Calculate the maximum height reached
Cancelling \(m\) from both sides, \[ 50=10h \] \[ h=5 \] Therefore, \[ \boxed{h=5\,\text{m}} \]

Step 6: Select the correct option
The maximum height attained by the bob is \[ \boxed{5\,\text{m}} \] Hence the correct answer is \[ \boxed{\text{Option (D)}} \]