Question

A 50 g ball of clay traveling at speed Vo hits and sticks to a 1.0 kg block sitting at rest on a frictionless surface. What is the speed of the block after the collision? What percentage of the ball's initial energy is "lost"?

Answer 1

Concepts and reason

The concepts used to solve this problem, are conservation of linear momentum and kinetic energy.

Initially, use the relation between the mass of the ball, the mass of the block, initial velocity of the ball, and the initial velocity of the block to calculate the velocity of the two objects just after the collision.

Equate the initial momentum before the collision and final momentum after the collision to calculate the velocity of the block after the collision.

Finally, find the percentage of energy lost by the balls in its initial kinetic energy.

Fundamentals

Conservation of linear momentum states that “the momentum of the two objects before collision is equal to the momentum of the two objects after collision”.

Both two objects collide together and remain together after the collision.

Apply the conservation of linear momentum to the system.

${m_1}{u_1} + {m_2}{u_2} = \left( {{m_1} + {m_2}} \right)v$

Here, ${m_1}$ is the mass of the ball, ${m_2}$ is the mass of the block, ${u_1}$ is the initial velocity of the ball before collision, ${u_2}$ is the initial velocity of the block before collision, and v is the velocity of the block after the collision.

Expression for the initial kinetic energy before the collision is,

$K.{E_{1i}} = \frac{1}{2}{m_1}u_1^2$

Here, $K.{E_{1i}}$ is the initial kinetic energy before the collision.

Expression for the final kinetic energy after the collision when the ball hits the block is,

$K.{E_{1f}} = \frac{1}{2}\left( {{m_1} + {m_2}} \right){v^2}$

Here, $K.{E_{1f}}$ is the final kinetic energy after the collision when the ball hits the block.

Expression for the percentage loss in the ball’s initial kinetic energy is,

$\left( {K \cdot {E_i}} \right)\% = \frac{{K.{E_{1i}} - K.{E_{1f}}}}{{K.{E_{1i}}}}$

Here, $\left( {K \cdot {E_i}} \right)\%$ is the percentage loss in the ball’s initial kinetic energy.

(1)

The conservation of linear momentum to the system is,

${m_1}{u_1} + {m_2}{u_2} = \left( {{m_1} + {m_2}} \right)v$

Rearrange the above equation to get v.

$v = \frac{{{m_1}{u_1} + {m_2}{u_2}}}{{\left( {{m_1} + {m_2}} \right)}}$

Substitute $50\,{\rm{g}}$ for ${m_1}$ and $1.0\,{\rm{kg}}$ for ${m_2}$, ${v_0}$ for ${u_1}$, and $0\,{\rm{m/s}}$ for ${u_2}$.

$\begin{array}{c}\\v = \frac{{\left( {50\,{\rm{g}}\left( {\frac{{{{10}^{ - 3}}\,{\rm{kg}}}}{{1\,{\rm{g}}}}} \right)} \right){v_0} + \left( {1.0\,{\rm{kg}}} \right)\left( {0\,{\rm{m/s}}} \right)}}{{\left( {50\,{\rm{g}}\left( {\frac{{{{10}^{ - 3}}\,{\rm{kg}}}}{{1\,{\rm{g}}}}} \right) + 1.0\,{\rm{kg}}} \right)}}\\\\ = \frac{{\left( {0.05\,{\rm{kg}}} \right){v_0} + 0\,{\rm{kg}} \cdot {\rm{m/s}}}}{{0.05\,{\rm{kg}} + 1.0\,{\rm{kg}}}}\\\\ = \left( {0.0476{v_0}} \right)\,{\rm{m/s}}\\\end{array}$

Here, ${v_0}$ is the initial velocity of the ball.

(2)

Expression for the percentage loss in ball’s initial kinetic energy is as follows:

$\left( {K \cdot {E_i}} \right)\% = \frac{{K.{E_{1i}} - K.{E_{1f}}}}{{K.{E_{1i}}}} \times 100$ …… (1)

Expression for the initial kinetic energy before the collision is,

$K.{E_{1i}} = \frac{1}{2}{m_1}u_1^2$ …… (2)

Expression for the final kinetic energy after the collision when the ball hits the block is,

$K.{E_{1f}} = \frac{1}{2}\left( {{m_1} + {m_2}} \right){v^2}$ …… (3)

Substituting equation (3) and (2) in the equation (1).

$\left( {K \cdot {E_i}} \right)\% = \frac{{\frac{1}{2}{m_1}u_1^2 - \frac{1}{2}\left( {{m_1} + {m_2}} \right){v^2}}}{{\frac{1}{2}{m_1}u_1^2}} \times 100$

Substitute $50\,{\rm{g}}$ for ${m_1}$ and $1.0\,{\rm{kg}}$ for ${m_2}$, ${v_0}$ for ${u_1}$, and $\left( {0.0476{v_0}} \right)\,{\rm{m/s}}$ for $v$.

$\begin{array}{c}\\\left( {K \cdot {E_i}} \right)\% = \frac{{\frac{1}{2}\left( {50\,{\rm{g}}\left( {\frac{{{{10}^{ - 3}}\,{\rm{kg}}}}{{1\,{\rm{g}}}}} \right)} \right){{\left( {{v_0}} \right)}^2} - \frac{1}{2}\left( {\left( {50\,{\rm{g}}\left( {\frac{{{{10}^{ - 3}}\,{\rm{kg}}}}{{1\,{\rm{g}}}}} \right)} \right) + 1.0\,{\rm{kg}}} \right){{\left[ {\left( {0.0476{v_0}} \right)\,{\rm{m/s}}} \right]}^2}}}{{\frac{1}{2}\left( {50\,{\rm{g}}\left( {\frac{{{{10}^{ - 3}}\,{\rm{kg}}}}{{1\,{\rm{g}}}}} \right)} \right){{\left( {{v_0}} \right)}^2}}} \times 100\\\\ = \frac{{\left( {0.05\,{\rm{kg}}} \right){{\left( {{v_0}} \right)}^2} - \left( {0.002379048} \right){{\left( {{v_0}} \right)}^2}}}{{\left( {0.05\,{\rm{kg}}} \right){{\left( {{v_0}} \right)}^2}}} \times 100\\\\ = 95.2\% \\\end{array}$

Ans: Part 1

Thus, the block’s speed after the collision is $\left( {0.0476{v_0}} \right)\,{\bf{m/s}}$.

Part 2

Thus, the percentage of loss in the ball’s initial kinetic energy is ${\bf{95}}{\bf{.2\% }}$.