Question

A block of mass $1.9\mathrm{kg}$ is at rest at the edge of a table, of height $1\mathrm{m}$. A bullet of mass $0.1\mathrm{kg}$ collides with the block and sticks to it. If the velocity of the bullet is $20\mathrm{m}/\mathrm{s}$ in the horizontal direction just before the collision then the kinetic energy just before the combined system strikes the floor, is [Take $\mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2$ Assume there is no rotational motion and loss of energy after the collision is negligible.]

• A. $23\mathrm{J}$
• B. $21\mathrm{J}$
• C. $20\mathrm{J}$
• D. $19\mathrm{J}$

Answer 1

The correct option is B

$21\mathrm{J}$

Step 1: Given data and diagram

Mass of block, ${\mathrm{m}}_1=1.9\mathrm{kg}$

Mass of bullet, ${\mathrm{m}}_2=0.1\mathrm{kg}$

Mass of combined system, $\mathrm{m}={\mathrm{m}}_1+{\mathrm{m}}_2=2\mathrm{kg}$

Velocity of bullet, $\mathrm{v}=20{\mathrm{ms}}^{-1}$

Let the velocity of the combined system$={\mathrm{V}}_{\mathrm{c}}$

The block is at rest on a table which is at a height of $\mathrm{h}=1\mathrm{m}$

The diagram for the given problem is shown below:

Step 2: Calculating the kinetic energy of the combined system

Apply law of conservation of linear momentum, we get

momentum before collision = momentum after collision

$$\begin{gathered} \Rightarrow \mathrm{m}\times \mathrm{v}=\left({\mathrm{m}}_1+{\mathrm{m}}_2\right){\mathrm{V}}_{\mathrm{c}} \\ \Rightarrow 0.1\times 20=2\times {\mathrm{V}}_{\mathrm{c}} \\ \Rightarrow {\mathrm{V}}_{\mathrm{c}}=1{\mathrm{ms}}^{-1} \end{gathered}$$

Now the kinetic energy of the system will be,

$$\begin{gathered} \mathrm{Kinetic}\mathrm{energy}=\frac{1}{2}{{\mathrm{mV}}^2}_{\mathrm{c}} \\ =\frac{1}{2}\times 2\times {\left(1\right)}^2 \\ =1\mathrm{J} \end{gathered}$$

And total energy of the system will be,

$$\begin{gathered} \mathrm{Total}\mathrm{energy}=\mathrm{kinetic}\mathrm{energy}+\mathrm{mgh} \\ =1+\left(2\times 10\times 1\right) \\ =21\mathrm{J} \end{gathered}$$

Therefore the kinetic energy is equal to $1\mathrm{J}$ and total energy before striking the ground is equal to $21\mathrm{J}$

Hence, option (B) is the correct answer.