a man 60 kg skating with a speed of 10 m/s collides with 40 kg skater at rest and they cling to eachother. Find the loss in kinetic energy?
figure initial KE
1/2 60 10^2=
final KE
momentum.. (60+40)V=60*10 V= 6m/s
final ke then 1/2 100*6^2
you figure the loss.
1800J
Bobpursley can u write it for me in a better way than what you have written
To find the loss in kinetic energy, we need to calculate the initial and final kinetic energy before and after the collision.
Let's start by calculating the initial kinetic energy of the first skater:
Initial kinetic energy (KE1) = 1/2 * mass1 * velocity1^2
= 1/2 * 60 kg * (10 m/s)^2
= 1/2 * 60 kg * 100 m^2/s^2
= 3000 kg*m^2/s^2
Now, let's calculate the initial kinetic energy of the second skater:
Initial kinetic energy (KE2) = 1/2 * mass2 * velocity2^2
= 1/2 * 40 kg * (0 m/s)^2
= 1/2 * 40 kg * 0 m^2/s^2
= 0 kg*m^2/s^2
After the collision, the two skaters cling to each other and move together as one. This means their final velocity will be the same. Let's call it V.
Using the principle of conservation of momentum, we can say:
(mass1 * velocity1) + (mass2 * velocity2) = (mass1 + mass2) * V
Substituting the given values:
(60 kg * 10 m/s) + (40 kg * 0 m/s) = (60 kg + 40 kg) * V
600 kg*m/s = 100 kg * V
V = 600 kg*m/s / 100 kg
V = 6 m/s
Now, let's calculate the final kinetic energy:
Final kinetic energy (KEf) = 1/2 * (mass1 + mass2) * V^2
= 1/2 * (60 kg + 40 kg) * (6 m/s)^2
= 1/2 * 100 kg * 36 m^2/s^2
= 1800 kg*m^2/s^2
To find the loss in kinetic energy, we subtract the final kinetic energy from the initial kinetic energy:
Loss in Kinetic Energy = Initial KE - Final KE
= 3000 kg*m^2/s^2 - 1800 kg*m^2/s^2
= 1200 kg*m^2/s^2
Therefore, the loss in kinetic energy is 1200 kg*m^2/s^2.