Question

A 75-g projectile traveling at 600 m /s strikes and becomes embedded in the 40-kg block, which is ini-tially stationary. Compute the energy lost during the impact. Express your answer as an absolute value ΔE| and as a percentage n of the original system energy E.

Answer 1

Answer:

Explanation:

Answer:

Explanation:

m1 = 75 g = 0.075 kg

m2 = 40 kg

u1 = 600 m/s

u2 = 0

Let the velocity of combined mass is v.

Use conservation of momentum

m1 x u1 + m2 x u2 = ( m1 + m2) x v

0.075 x 600 + 0 = ( 40 + 0.075) x v

v = 1.12 m/s

Initial energy, E = 0.5 x 0.075 x 600 x 600 = 13500 J

Final energy, E' = 0.5 x 40.075 x 1.12 x 1.12 = 25.14 J

Loss in energy, ΔE = E - E' = 13500 - 25.14 = 13474.86 J

% loss of energy = ( 13474.86 x 100) / 13500 = 99.8 %  

Answer 2

Explanation:

The given data is as follows.

          Mass, m = 75 g

         Velocity, v = 600 m/s

As no external force is acting on the system in the horizontal line of motion. So, the equation will be as follows.

          $m_{1}v_{1_{i}} = (m_{1} + m_{2})vi$

where,  $m_{1}$ = mass of the projectile

            $m_{2}$ = mass of block

              v = velocity after the impact

Now, putting the given values into the above formula as follows.

              $m_{1}v_{1_{i}} = (m_{1} + m_{2})vi$

         $75(10^{-3}) \times 600 = [(75 \times 10^{-3}) + 50] \times v$

                                  = $\frac{45}{50.075}$

                              v = 0.898 m/s

Now, equation for energy is as follows.

               E = $\frac{1}{2}mv^{2}$

                  = $\frac{1}{2} \times (75 \times 10^{-3} + 50) \times (600)^{2}$

                  = 13500 J

Now, energy after the impact will be as follows.

             E' = $\frac{1}{2}[75 \times 10^{-3} + 50](0.9)^{2}$

                 = 20.19 J

Therefore, energy lost will be calculated as follows.

           $\Delta E$ = E  E'

                       = (13500 - 20) J

                       = 13480 J

And,   n = $\frac{\Delta E}{E}$

             = $\frac{13480}{13500} \times 100$

             = 99.85

             = 99.9%

Thus, we can conclude that percentage n of the original system energy E is 99.9%.