Question

Two balls, $X$, and $Y$ move along a horizontal frictionless surface, as illustrated in fig. Ball $X$ has an initial velocity $3.0m{s^{ - 1}}$in a direction along line $AB$. Ball $Y$has a mass of $2.5kg$and an initial velocity $9.6m/{s^{ - 1}}$in the direction at an angle of ${60^ \circ }$ to line $AB$.
The two balls collide at the point $B$. The balls stick together and then travel along the horizontal surface in a direction at right-angles to the line $AB$, as shown in Fig. Determine the difference between the initial kinetic energy of the ball $X$ and the initial kinetic energy of the ball $Y$.
The difference in kinetic energy=________________$J$.

Answer 1

Hint We have two balls moving on a frictionless horizontal surface. The masses and velocities of the two balls are given. If the two balls collide at a given point the balls will stick together as shown in the figure and move in a direction perpendicular to the horizontal surface. We have to find the difference in kinetic energy.
Formula used
$KE = \dfrac{1}{2}m{v^2}$
Where $KE$ is the kinetic energy, $m$ stands for the mass of the body, and $v$ stands for the velocity of the body.

Complete Step by step solution
To find the difference in kinetic energy, first, we have to find the mass of the ball $X$
The initial velocity of $X = 3.0m/s$
The initial velocity of $Y = 9.6m/s$
The angle between$A$and $B$ are given as.${60^ \circ }$
By applying the force on a free body diagram
After the collision, the net force will be zero
i.e. $\sum {F = 0}$
This means that the change in momentum will be zero
We can write the change in momentum as,
${m_x}{v_x} - {m_y}{v_y}$ $\left( {\because P = mv} \right)$
Where ${m_x}$,${m_y}$and, ${v_x}$ ${v_y}$are the masses and velocities of the two balls $X$and $Y$respectively.
${v_y}\cos \theta$is the velocity component of the ball $Y$in the $x$direction.
${v_x} = 3.0m/s$
${m_y} = 2.5kg$
${v_y} = 9.6m/s$
$\theta = {60^ \circ }$
Substituting the values in the above equation, we get ${m_x}$as
$(m \times 3.0) - (2.5 \times 9.6 \times \cos {60^ \circ }) = 0$
From this,
$3m = 12$
$\Rightarrow m = \dfrac{{12}}{3}=4kg$
Now that we have the masses and velocities of the two balls we can find the difference in kinetic energies.
We know that the kinetic energy of a body is,
$KE = \dfrac{1}{2}m{v^2}$
The difference in kinetic energies can be written as,
$\Delta KE = \dfrac{1}{2}{m_2}v_2^2 - \dfrac{1}{2}{m_1}v_1^2$
Substituting the values of masses and velocities, we get
$\Delta KE = \left( {\dfrac{1}{2} \times 2.5 \times {{9.6}^2}} \right) - \left( {\dfrac{1}{2} \times 4 \times {{3.0}^2}} \right)$
$\Delta KE = \dfrac{{230.4}}{2} - \dfrac{{36}}{2}$
$\Rightarrow \Delta KE = 115.2 - 18 = 97.2$

Therefore, the difference in kinetic energy $= 97.2J$

Note
Collisions are of two types. If the total kinetic energy of the system is conserved after the collision such collisions can be called elastic collisions. If the kinetic energy of the system is not conserved after collision such collisions are called inelastic collisions. The total linear momentum of the system will be conserved in all collisions.