Initial conditions:
m1 = 1.0 ; v1 = 5
m2 = 4.0 ; v2 = 0
In the case where the second object (sometimes called the target) is at rest the velocities after the condition are
v1' = v1* (m1-m2)/(m1+m2)
v2' = 2v1*m1/(m1+m2)
For this we get
v1' = 5*(-3)/5 = -3m/s (moving in the opposite direction as before at 3m/s
v2' = 2*5*(1)/5 = 2m/s in the same direction as the original ball was moving
you can see these directions by looking at the signs. The momenta also add to the initial momentum as required.
Unit it is c trust please
Rock is completely immersed in hot water. By the second law of thermodynamics, thermal energy or heat is transferred from substance with higher temperature to substance with lower temperature until they come to thermal equilibrium i.e. both at same temperature.
It is given here that rock is at 20°C which is at lower temperature than water at 80°C. ∴Heat or thermal energy flows from water to rock. So, right choice is-
A. The water gives the rock thermal energy and gets no thermal energy in return.
