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

Step 1: List the collision data.
Particle $A$ of mass $m$ moves at $u_A = 10\,\text{m s}^{-1}$ and strikes bob $B$ (also mass $m$, at rest) elastically. String length is $10\,\text{m}$, $g = 10\,\text{m s}^{-2}$.
Step 2: Use the equal-mass elastic rule.
In a head-on elastic collision between equal masses, the two particles exchange velocities.
Step 3: Velocities after impact.
So $A$ stops and $B$ moves off with the incoming speed.
\[ v_A = 0, \qquad v_B = 10\,\text{m s}^{-1} \]
Step 4: Kinetic energy given to the bob.
\[ K = \tfrac12 m v_B^2 = \tfrac12 m (10)^2 = 50m \]
Step 5: Convert to height by energy conservation.
As the bob swings up, its kinetic energy becomes gravitational potential energy $mgh$.
\[ \tfrac12 m v_B^2 = mgh \Rightarrow 50m = m(10)h \]
Step 6: Solve for the height.
Cancel $m$: $50 = 10h$, so $h = 5\,\text{m}$.
\[ \boxed{h = 5\,\text{m}} \]