Question
Question: A trolley of mass \(60\,kg\) moves on a smooth horizontal surface and has kinetic energy \(120\,J\)....
A trolley of mass 60kg moves on a smooth horizontal surface and has kinetic energy 120J. A mass of 40kg is lowered vertically onto the trolley. The total kinetic energy of the system after lowering the mass is:
A) 60J
B) 72J
C) 120J
D) 144J
Solution
The kinetic energy depends upon mass and velocity of the body. The total mass will be the sum of mass of the trolley and the mass that is lowered onto the trolley. To find the velocity of the trolley after lowering the mass we can use the conservation of momentum. According to conservation of momentum the initial momentum will be equal to the final momentum.
Complete step by step solution:
It is given that a trolley has a mass, m1=60kg .
The kinetic energy of the trolley is
KE=120J .
We need to find the kinetic energy when a mass of 40kg is lowered vertically onto the trolley.
The kinetic energy is the energy possessed by a body by virtue of its motion. Kinetic energy can be calculated as
KE=21mv2
Where, m is the mass, v is the velocity.
By substituting the given values in the equation, we can find the value of initial velocity, vi as
⇒120=21×60×vi2
⇒vi=2m/s
In order to find the final velocity after lowering the mass onto the trolley we can use the law of conservation of momentum.
According to conservation of momentum the final momentum will be equal to the initial momentum.
Momentum is calculated as a product of mass and velocity.
⇒P=mv.
Where, m is the mass and v is the velocity.
Hence initial momentum is given as
⇒Pi=mivi
Final momentum is given as
⇒Pf=mfvf
Now let us equate both initial and final momentum.
⇒mfvf=mivi
The final mass is
mf=mi+40
⇒mf=60+40=100
On substituting the values, we get
⇒100×vf=60×2
⇒vf=1.2m/s
This is the value of final velocity.
Thus, the final kinetic energy can be calculated as
KEf=21mfvf2
On substituting the values, we get
⇒KEf=21×100×(1.2)2
∴KEf=72J
This is the final kinetic energy of the system.
So, the correct answer is option B.
Note: We can apply the momentum conservation only when the net force acting on the system is zero. Here, since the trolley is moving on a smooth horizontal surface there is no effect of frictional force. When friction is also taken into consideration, we cannot apply the conservation of linear momentum.