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

![](https://res.cloudinary.com/ddr1kuyb3/image/upload/v1786186759/questions/nqrzpbdwximsjiuno1rd.png)

Question

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

![](https://res.cloudinary.com/ddr1kuyb3/image/upload/v1786186759/questions/nqrzpbdwximsjiuno1rd.png)

Accepted Answer

D. 1. Notice that mass $A$ and bob $B$ have **equal mass** $m$, and they collide **elastically** in one dimension.
Recall the key result for an elastic collision between two equal masses: the moving mass comes to rest, and its entire velocity is transferred to the stationary mass.
Hence, after the collision, bob $B$ moves off with the full initial speed of $A$.
$$v_B=10\ \mathrm{ms}^{-1}$$

2. Since the bob is attached to a string and swings upward like a pendulum, use **conservation of energy** to relate its speed just after collision to the height $h$ it rises.
Given that all kinetic energy converts into gravitational potential energy at the highest point,
$$mgh=\frac{1}{2}mv_B^2$$

3. Cancel the mass $m$ from both sides and solve for $h$.
$$h=\frac{v_B^2}{2g}$$

4. Substitute $v_B=10\ \mathrm{ms}^{-1}$ and $g=10\ \mathrm{ms}^{-2}$.
$$h=\frac{(10)^2}{2	imes10}=\frac{100}{20}=5\ \mathrm{m}$$

Hence, the bob rises to a height of 5 m, matching **Option D**.