13. A 60 kg boy lying on a surface of negligiblefriction throws horizontally a stone of mass 1kg with a speed of 12 m/s away from him. Asa result with what kinetic energy he moves back
Answers
Answer:
1.2 J
Explanation:
Momentum gained by stone = 1*12
= 12 kgm/s
The speed gained by boy = 12/60
= 0.2m/s
Kinetic Energy gained = (1/2)(m)v^2
= 0.5(60)(0.2)^2
= 30*0.2*0.2
= 1.2 J
Answer:
Explanation:
The relevant principle is that momentum is conserved. Therefore, as momentum is just velocity x mass, then the stone has gain 1 x 12 = 12 kgm/s. As momentum is conserved (and friction is negligible), then that means the boy gains 12 kgm/s in the opposite direction. As the boy has a mass of 60 kg, then this means his gain in velocity will be 12 / 60 = 0.2 m/s.
If we assume that the boy was at rest before the stone was thrown, then we just have to plug the numbers into the equation for kinetic energy, that is Ke=12mv2. This means that the boy will be travelling in the opposite direction to the stone at 0.2 m/s with kinetic energy of 1.2 J.