Chapter 7, Problem 78GP

(a)

To determine

The height up to which the skeet will go.

Answer to Problem 78GP

The pellet and skeet system goes 6.5 m vertically higher.

Explanation of Solution

Given:

The skeet of mass $M=0.25\text{kg}$ is projectile at an angle of $30°$ with the horizontal at a speed of $25 \text{m/s}$ . At its highest vertical position, it is hit by a pallet of mass $m=15 \text{g}$ which is moving in vertically straight motion with a speed of $200 \text{m/s}$ and sticks with the skeet. Final velocity of the pallet is $v_y=0$ at the top.

Formula used:

According to the conservation of momentum principle, the total momentum before and after collision remains constant $p_i=p_f$

Newton’s third equation of motion is $v^2=v_0^2+2as$

The range of the projectile motion is $R=\frac{v_0^{2}sin2\theta }{g}$

Calculation:

Range of the projectile motion is

  $\begin{array}{l}R=\frac{v_0^{2}sin2\theta }{g}\\ R=\frac{{\left(25\text{m/s}\right)}^{2}sin60°}{(9.81\text{m/s})}=55.2\text{m}\end{array}$

Considering vertically upward being positive. Now, using Newton’s second equation of motion to find the vertical distance traveled

  $\begin{array}{c}{v}^2=v_0^2+2as\\ s=\frac{v^2-v_0^2}{2a}\\ =\frac{0-{[(25)sin30°]}^2}{2\times \left(-9.8\right)}\\ s=7.97\text{m}\end{array}$

Now, calculating the horizontal and vertical component of velocity after the elastic collision is

  $\begin{array}{c}M{v}_x=(m+M)v_x^{\text{'}}\\ v_x^{\text{'}}=\frac{M{v}_x}{m+M}\\ =\frac{\text{(0}{\text{.25 kg) (25m/s)cos30}}^{\text{o}}}{\text{(0}\text{.25+0}\text{.015)kg}}\\ v_x^{\text{'}}=20.42\text{m/s}\end{array}$ ;

  $\begin{array}{l}m{v}_y=(m+M)v_y^{\text{'}}\\ v_y^{\text{'}}=\frac{m{v}_y}{m+M}=\frac{\text{(0}\text{.015 kg) (200m/s)}}{\text{(0}\text{.25+0}\text{.015)kg}}\\ v_y^{\text{'}}=11.32\text{m/s}\end{array}$

Now, calculating extra height is achieved by the system of pallet and skeet using vertical speed.

  $\begin{array}{l}{v}^2=v_0^2+2a{s}_{extra}\\ s_{extra}=\frac{v^2-v_0^2}{2a}=\frac{\text{0}-{\text{(11}\text{.32m/s)}}^{\text{2}}}{\text{2×}\left(-\text{9}{\text{.8m/s}}^{\text{2}}\right)}\\ s_{extra}=6.5\text{m}\end{array}$

Conclusion:

The pellet and skeet system goes 6.5m vertically higher.

(b)

To determine

The extra distance $\Delta x$ , that the skeet will travel because of the collision.

Answer to Problem 78GP

The extra distance $\Delta x$ traveled by the skeet because of the collision is 31.2 m.

Explanation of Solution

Given:

The skeet of mass $M=0.25\text{kg}$ is projectile at an angle of $30°$ with the horizontal at a speed of $25 \text{m/s}$ . At its highest vertical position, it is hit by a pallet of mass $m=15 \text{g}$ which is moving in vertically straight motion with a speed of $200 \text{m/s}$ , and sticks with the skeet. Final velocity of the pallet is $v_y=0$ at the top.

Formula used:

Newton’s second equation of motion is $s=ut+\frac{1}{2}a{t}^2$

Distance-speed relationship is $s=vt$

Calculation:

The time required to fall to the ground would be

  $\begin{array}{l}y=y_0+v_y^{\text{'}}t+\frac{1}{2}a{t}^2\\ 0=7.97 +(11.32 )t+\frac{1}{2}(-9.8 )t^2\\ 4.9t^2-11.32t-7.97=0\\ t=2.88 \sec , -0.565 \sec \end{array}$

Since time cannot be negative. The accepted value of time would be $s=vt$

Thus, horizontal distance traveled by the system after the collision is

  $\begin{array}{l}{x}_{after}=v_x^{\text{'}}t\\ x_{after}=(20.42)(2.88)\\ x_{after}=58.8\text{m}\end{array}$

Horizontal distance traveled by the skeet in the non-collision scenario would be $\frac{1}{2}R$ . Thus, extra distance traveled due to collision is

  $\begin{array}{l}\Delta x=x_{after}-\frac{1}{2}R\\ \Delta x=58.8-\frac{1}{2}(55.2)=31.2\text{m}\end{array}$

Conclusion:

The extra distance $\Delta x$ traveled by the skeet because of the collision is 31.2 m