A)Ep'=mgh=mgl(1-cosa).At the bottom of the swing Ep=0(reference level),so the potential energy as the child is just released is bigger than the potential energy at the bottom of the swing.;B)The speed of the child at the bottom of the swing-->v=√(2gh)=√[2gl(1-cosa)];C)I don't think that the tension does any work.
<span>C. It is the difference in electrical potential energy between two places in an electric field.</span>
The momentum of the second ball was 15 kg.m/s.
<h3>What is inelastic collision?</h3>
In which collision some amount of kinetic energy of the system is lost that called inelastic collision. In purely inelastic collision, two bodies stick together. But principle of conservation of linear momentum is obeyed.
In the given question,
Two balls collide and after collision, the final momentum of the system = 18 kg.m/s.
Initial velocity of 1st ball of mass 3 kg is 1 m/s.
So, Initial momentum of first ball = mass × velocity = (3 kg) × (1 m/s) = 3 kg.m/s.
According to Principle of conservation of linear momentum for this inelastic collision,
Initial momentum of first ball + initial momentum of second ball = final momentum of the system
⇒ initial momentum of second ball = final momentum of the system - Initial momentum of first ball
= 18 kg.m/s - 3 kg.m/s.
= 15 kg.m/s.
Hence, initial momentum of second ball = 15 kg.m/s.
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