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Kinematics in 2D is about the motion of an object in a plane. Three vectors represent the path. The motion of the object is separated into two components- horizontal (along the x-axis) and vertical (along the y-axis) components. Kinematic equations describe the individual motions.

The principle of conservation of linear momentum states that if two bodies collide with each other, the total linear momentum before and after the collision remains the same if no external force acts on the system.

\(\begin{array}{c}{p_{{\rm{before}}}} = {p_{{\rm{after}}}}\\{\left( {mv} \right)_{{\rm{before}}}} = {\left( {mv} \right)_{{\rm{after}}}}\end{array}\) 鈥� (i)

The mass of the skeet is\(M = 0.25\;{\rm{kg}}\).

The angle of projection is\(\theta = {28^{\rm{o}}}\).

The launching speed of the skeet is\({v_0} = 25\;{\rm{m/s}}\).

The mass of the pellet is\(m = 15\;{\rm{g}} = 0.015\;{\rm{kg}}\).

The launching speed of the pellet is\(v = 230\;{\rm{m/s}}\).

The original horizontal distance or the range of the skeet is given by:

\(\begin{array}{c}R = \frac{{v_0^2\sin 2\theta }}{g}\\ = \frac{{{{\left( {25\;{\rm{m/s}}} \right)}^2}\sin {{56}^{\rm{o}}}}}{{9.8\;{\rm{m/}}{{\rm{s}}^2}}}\\ = 52.87\;{\rm{m}}\end{array}\)

The height at which the skeet and the pellet collide can be found using the following kinematic equation of motion. Also, note that at the top, the final velocity is zero\(\left( {v = 0} \right)\).

\(\begin{array}{c}{v^2} - {u^2} = 2a\Delta y\\0 - {\left( {{v_0}\sin \theta } \right)^2} = 2\left( { - g} \right)\Delta y\\\Delta y = \frac{{ - {{\left( {{v_0}\sin \theta } \right)}^2}}}{{2\left( { - g} \right)}}\end{array}\)

Substituting the numerical values in the above expression, you get:

\(\begin{array}{c}\Delta y = \frac{{ - {{\left( {\left( {25\;{\rm{m/s}}} \right)\left( {\sin {{28}^{\rm{o}}}} \right)} \right)}^2}}}{{2\left( { - 9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)}}\\ = 7.028\;{\rm{m}}\end{array}\)

After the collision, the skeet moves horizontally at the speed:

\(\begin{array}{c}{v_{\rm{x}}} = {v_0}\cos \theta \\ = \left( {25\;{\rm{m/s}}} \right)\cos {28^{\rm{o}}}\\ = 22.07\;{\rm{m/s}}\end{array}\)

The vertical velocity of the bullet is\({v_{\rm{y}}} = 230\;{\rm{m/s}}\).

Apply the conservation of momentum along the x-direction.

\(\begin{array}{c}M{v_{\rm{x}}} = \left( {m + M} \right){v_{\rm{x}}}^\prime \\{v_{\rm{x}}}^\prime = \frac{{M{v_{\rm{x}}}}}{{\left( {m + M} \right)}}\\ = \frac{{\left( {0.25\;{\rm{kg}}} \right)\left( {22.07\;{\rm{m/s}}} \right)}}{{\left( {0.25\;{\rm{kg}} + 0.015\;{\rm{kg}}} \right)}}\\ = 20.82\;{\rm{m/s}}\end{array}\)

The final velocity of the two objects after an inelastic collision along the x-direction is\(20.82\;{\rm{m/s}}\).

Apply the conservation of momentum along the y-direction.

\(\begin{array}{c}m{v_{\rm{y}}} = \left( {m + M} \right){v_{\rm{y}}}^\prime \\{v_{\rm{y}}}^\prime = \frac{{m{v_{\rm{y}}}}}{{\left( {m + M} \right)}}\\ = \frac{{\left( {0.015\;{\rm{kg}}} \right)\left( {230\;{\rm{m/s}}} \right)}}{{\left( {0.25\;{\rm{kg}} + 0.015\;{\rm{kg}}} \right)}}\\ = 13.02\;{\rm{m/s}}\end{array}\)

The final velocity of the two objects after the inelastic collision along the y-direction is \(13.02\;{\rm{m/s}}\).

The height h鈥� is the height to which the skeet-pellet combination rises above the point of collision. Using the kinematic equation of motion, you get:

\(\begin{array}{c}v_{\rm{y}}^2 - v_{{\rm{y0}}}^2 = 2ah'\\h' = \frac{{v_{\rm{y}}^2 - v_{{\rm{y0}}}^2}}{{2\left( { - g} \right)}}\\ = \frac{{0 - {{\left( {13.02\;{\rm{m/s}}} \right)}^2}}}{{2\left( { - 9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)}}\\ = 8.65\;{\rm{m}}\end{array}\)

Using the following kinematic equation of motion, you get:

\(\begin{array}{c}h' = {v_{0{\rm{y}}}}t + \frac{1}{2}g{t^2}\\ - 8.65\;{\rm{m}} = \left( {13.02\;{\rm{m/s}}} \right)t + \frac{1}{2}\left( { - 9.8\;{\rm{m/}}{{\rm{s}}^2}} \right){t^2}\\0 = 4.9{t^2} - 13.02t - 8.65\\t = 3.207\;{\rm{s,}}\; - {\rm{0}}{\rm{.550}}\;{\rm{s}}\end{array}\)

Neglect the negative value of time taken and consider \(t = 3.207\;{\rm{s}}\).

The horizontal distance traveled by the combination of skeet and pellet is given by:

\(\begin{array}{c}{x_{{\rm{after}}}} = {v_{\rm{x}}}^\prime t\\ = \left( {20.82\;{\rm{m/s}}} \right)\left( {3.207\;{\rm{s}}} \right)\\ = 66.77\;{\rm{m}}\end{array}\)

If the collision between the skeet and the pellet had not taken place, the skeet would have covered a horizontal distance equal to\(\frac{1}{2}R\), where R is the range of the skeet.

Therefore, the extra distance\(\Delta x\)traveled by the skeet after the collision is given by:

\(\begin{array}{c}\Delta x = {x_{{\rm{after}}}} - \frac{1}{2}R\\ = 66.77\;{\rm{m}} - \frac{1}{2}\left( {52.87\;{\rm{m}}} \right)\\ = 40.33\;{\rm{m}}\end{array}\)