a 70 kg man standing on ice throws a 3 kg body horizontally at 8 m/s. the friction coefficient between the ice and his feet is 0
1 answer:
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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.
where, = mass of the projectile
= mass of block
v = velocity after the impact
Now, putting the given values into the above formula as follows.
=
v = 0.898 m/s
Now, equation for energy is as follows.
E =
=
= 13500 J
Now, energy after the impact will be as follows.
E' =
= 20.19 J
Therefore, energy lost will be calculated as follows.
= E E'
= (13500 - 20) J
= 13480 J
And, n =
=
= 99.85
= 99.9%
Thus, we can conclude that percentage n of the original system energy E is 99.9%.