Question

A bullet of mass $\mathbf{0}\mathbf{.}\mathbf{105}\mathbf{}\mathbf{kg}$traveling horizontally at a speed of role="math" localid="1658149679010" $\mathbf{300}\mathbf{}\raisebox{1ex}{$\mathbf{m}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{s}$}\right.$embeds itself in a block of mass $\mathbf{2}\mathbf{}\mathbf{kg}$that is sitting at rest on a nearly frictionless surface. What is the speed of the block after the bullet embeds itself in the block?

Short answer

The speed of the block after the bullet embeds itself in the block is $14.96\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$.

Step-by-step solution

Identification of the given data

The given data can be listed below as-

• The mass of the bullet is $0.105\mathrm{kg}$.

• The velocity of the bullet is $300\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$.

• The mass of the block is $2\mathrm{kg}$.

• The speed of the block is $0\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$as it is at rest.

Significance of the law of conservation of momentum for the block

This law states that an object’s momentum before and after the collision becomes equal if there are no external forces involved.

The equation of the velocity gives the speed of the block.

Determination of the speed of the block

From the law of conservation of momentum, the equation of the speed of the block is expressed as:

$\mathrm{V}=\frac{{\mathrm{m}}_1{\mathrm{v}}_1+{\mathrm{m}}_2{\mathrm{v}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}$

Here, v is the speed of the block, ${\mathrm{m}}_1$and ${\mathrm{m}}_2$are the mass of the bullet and the block and ${\mathrm{v}}_1$and ${\mathrm{v}}_2$are the velocities of the bullet and the block respectively.

Substituting the values in the above equation, we get-

$\mathrm{v}=\frac{(0.105\mathrm{kg})\times (300\mathrm{m}/\mathrm{s})+\left(2\mathrm{kg}\right)\times (0\mathrm{m}/\mathrm{s})}{0.105\mathrm{kg}+2\mathrm{kg}}$

$\mathrm{v}=14.96\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$

Thus, the speed of the block after the bullet embeds itself in the block is $14.96\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.$.