A 40-kg skater travelling at 4m/s overtakes a 60-kg skater traveling at 2m/s in the same direction and collides with him.
a. If two skater remains in contact, what is their final velocity?
b. How much kinetic energy is lost?
(a) conserve momentum:
40*4 + 60*2 = (40+60)v
(b) calculate KE = 1/2 mv^2 before and after the collision
To solve this problem, we can use the principle of conservation of momentum and the concept of kinetic energy.
a. To find the final velocity when the skaters remain in contact, we can use the equation for conservation of momentum:
(m1 * v1) + (m2 * v2) = (m1 + m2) * vf
where:
m1 = mass of the first skater = 40 kg
v1 = initial velocity of the first skater = 4 m/s
m2 = mass of the second skater = 60 kg
v2 = initial velocity of the second skater = 2 m/s
vf = final velocity of the two skaters
Plugging in the values into the equation, we get:
(40 kg * 4 m/s) + (60 kg * 2 m/s) = (40 kg + 60 kg) * vf
Simplifying the equation gives:
(160 kg⋅m/s) + (120 kg⋅m/s) = (100 kg) * vf
280 kg⋅m/s = 100 kg * vf
Dividing both sides of the equation by 100 kg, we get:
vf = 280 kg⋅m/s / 100 kg
vf = 2.8 m/s
Therefore, their final velocity when they remain in contact after the collision is 2.8 m/s.
b. The kinetic energy lost in the collision can be calculated by finding the initial kinetic energy of the system and subtracting the final kinetic energy.
The initial kinetic energy of the system is given by:
KE_initial = (1/2) * m1 * v1^2 + (1/2) * m2 * v2^2
Plugging in the values, we have:
KE_initial = (1/2) * 40 kg * (4 m/s)^2 + (1/2) * 60 kg * (2 m/s)^2
KE_initial = 320 J + 120 J
KE_initial = 440 J
The final kinetic energy of the system is given by:
KE_final = (1/2) * (m1 + m2) * vf^2
Plugging in the values, we get:
KE_final = (1/2) * (40 kg + 60 kg) * (2.8 m/s)^2
KE_final = (1/2) * 100 kg * 7.84 m^2/s^2
KE_final = 392 J
The kinetic energy lost during the collision is:
KE_lost = KE_initial - KE_final
KE_lost = 440 J - 392 J
KE_lost = 48 J
Therefore, the amount of kinetic energy lost in the collision is 48 J.