Respuesta :
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.
[tex]m_{1}v_{1_{i}} = (m_{1} + m_{2})vi[/tex]
where, [tex]m_{1}[/tex] = mass of the projectile
[tex]m_{2}[/tex] = mass of block
v = velocity after the impact
Now, putting the given values into the above formula as follows.
[tex]m_{1}v_{1_{i}} = (m_{1} + m_{2})vi[/tex]
[tex]75(10^{-3}) \times 600 = [(75 \times 10^{-3}) + 50] \times v[/tex]
= [tex]\frac{45}{50.075}[/tex]
v = 0.898 m/s
Now, equation for energy is as follows.
E = [tex]\frac{1}{2}mv^{2}[/tex]
= [tex]\frac{1}{2} \times (75 \times 10^{-3} + 50) \times (600)^{2}[/tex]
= 13500 J
Now, energy after the impact will be as follows.
E' = [tex]\frac{1}{2}[75 \times 10^{-3} + 50](0.9)^{2}[/tex]
= 20.19 J
Therefore, energy lost will be calculated as follows.
[tex]\Delta E[/tex] = E E'
= (13500 - 20) J
= 13480 J
And, n = [tex]\frac{\Delta E}{E}[/tex]
= [tex]\frac{13480}{13500} \times 100[/tex]
= 99.85
= 99.9%
Thus, we can conclude that percentage n of the original system energy E is 99.9%.
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 %
