A metal ball of mass 2Kg moving with speed of 36Km/h has a head on collision with a stationery ball of mass 3Kg. if after collision, both the balls move as a single mass, then the loss of K.E due to collision is:
Question
A metal ball of mass 2Kg moving with speed of 36Km/h has a head on collision with a stationery ball of mass 3Kg. if after collision, both the balls move as a single mass, then the loss of K.E due to collision is:
Solution
To solve this problem, we need to follow these steps:
Step 1: Convert the speed from km/h to m/s The speed of the metal ball is given as 36 km/h. We can convert this to m/s by multiplying by 1000 (to convert km to m) and dividing by 3600 (to convert hours to seconds). So, 36 km/h = 36 * 1000 / 3600 = 10 m/s.
Step 2: Calculate the initial kinetic energy The kinetic energy (KE) of an object is given by the formula KE = 1/2 * m * v^2, where m is the mass and v is the velocity. The initial kinetic energy of the metal ball is therefore KE_initial = 1/2 * 2 kg * (10 m/s)^2 = 100 J.
Step 3: Calculate the final kinetic energy After the collision, the two balls move together as a single mass. The total mass is now 2 kg + 3 kg = 5 kg. The final velocity of the combined mass is the initial momentum divided by the final mass, which is (2 kg * 10 m/s) / 5 kg = 4 m/s. The final kinetic energy is therefore KE_final = 1/2 * 5 kg * (4 m/s)^2 = 40 J.
Step 4: Calculate the loss in kinetic energy The loss in kinetic energy due to the collision is the initial kinetic energy minus the final kinetic energy. So, the loss is KE_initial - KE_final = 100 J - 40 J = 60 J.
So, the loss of kinetic energy due to the collision is 60 Joules.
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