A 3kg mass sliding on a frictionless surface has a velocity of 5m/s east when it undergoes a one-dimensional inelastic collision with a 2kg mass that has an initial velocity of 2m/s west. After the collision the 3kg mass has a velocity of 1m/s east. How much kinetic energy does the two-mass system lose during the collision?
Given:
M1 = 3kg, V1 = 5m/s.
M2 = 2kg, V2 = -2m/s.
V3 = Velocity of M1 after collision.
V4 = 1m/s = Velocity of M2 after collision.
Momentum before = Momentum after.
M1*V1 + M2*V2 = M1*V3 + M2*1.
3*5 + 2*(-2) = 3*V3 + 2*1,
V3 = 3 m/s.
KEb = 0.5*3*5^2 - 0.5*2*2^2 . = KE before collision.
KEa = 0.5*3*3^2 + 0.5*2*1^2. = KE after collision.
KE lost = KEb - KEa.
k
ea
To find the change in kinetic energy during the collision, we need to calculate the initial and final kinetic energy of the system.
The initial kinetic energy of the system can be calculated by summing the kinetic energies of the two masses before the collision:
KE_initial = (1/2) * m1 * v1^2 + (1/2) * m2 * v2^2
where m1 = 3 kg (mass of the 3 kg mass), m2 = 2 kg (mass of the 2 kg mass), v1 = 5 m/s (initial velocity of the 3 kg mass), and v2 = -2 m/s (initial velocity of the 2 kg mass). Since the 2 kg mass is moving west, its velocity is considered negative.
KE_initial = (1/2) * 3 kg * (5 m/s)^2 + (1/2) * 2 kg * (-2 m/s)^2
= (1/2) * 3 kg * 25 m^2/s^2 + (1/2) * 2 kg * 4 m^2/s^2
= 37.5 J + 4 J
= 41.5 J
The final kinetic energy of the system can be calculated by summing the kinetic energies of the two masses after the collision:
KE_final = (1/2) * m1 * v1'^2 + (1/2) * m2 * v2'^2
where v1' = 1 m/s (final velocity of the 3 kg mass) and v2' is the final velocity of the 2 kg mass. Since the system is one-dimensional and assumed to be inelastic, the two masses move with the same velocity after the collision.
v2' = v1' = 1 m/s
KE_final = (1/2) * 3 kg * (1 m/s)^2 + (1/2) * 2 kg * (1 m/s)^2
= (1/2) * 3 kg * 1 m^2/s^2 + (1/2) * 2 kg * 1 m^2/s^2
= 1.5 J + 1 J
= 2.5 J
The change in kinetic energy during the collision can be calculated by subtracting the final kinetic energy from the initial kinetic energy:
Change in KE = KE_initial - KE_final
= 41.5 J - 2.5 J
= 39 J
Therefore, the two-mass system loses 39 Joules of kinetic energy during the collision.
To determine the amount of kinetic energy lost during the collision, we need to compare the initial total kinetic energy of the system to the final total kinetic energy.
The initial total kinetic energy can be calculated by adding up the kinetic energies of the two masses before the collision:
Kinetic energy of mass 1: K1 = (1/2) * m1 * v1^2
where m1 is the mass of the 3kg mass and v1 is its initial velocity (5m/s east).
Kinetic energy of mass 2: K2 = (1/2) * m2 * v2^2
where m2 is the mass of the 2kg mass and v2 is its initial velocity (2m/s west).
The final total kinetic energy is calculated similarly, but using the final velocities of the masses after the collision:
Kinetic energy of mass 1 after collision: K1' = (1/2) * m1 * v1'^2
where v1' is the final velocity of the 3kg mass (1m/s east) after the collision.
Kinetic energy of mass 2 after collision: K2' = (1/2) * m2 * v2'^2
where v2' is the final velocity of the 2kg mass after the collision.
The kinetic energy lost during the collision is given by the difference between the initial and final total kinetic energies:
Kinetic energy lost = (K1 + K2) - (K1' + K2')
Now let's calculate the values:
K1 = (1/2) * 3kg * (5m/s)^2 = 37.5 J (joules)
K2 = (1/2) * 2kg * (2m/s)^2 = 4 J (joules)
K1' = (1/2) * 3kg * (1m/s)^2 = 1.5 J (joules)
To find the value of K2', we need to use the principle of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision. Since the collision is inelastic, the masses stick together after the collision.
Initially, the total momentum is given by the sum of the momentum of the two masses:
Initial momentum = m1 * v1 + m2 * v2
Final momentum is given by the momentum of the combined masses after the collision:
Final momentum = (m1 + m2) * v'
Using these equations, we can find the value of v':
Initial momentum = Final momentum
m1 * v1 + m2 * v2 = (m1 + m2) * v'
3kg * 5m/s + 2kg * (-2m/s) = (3kg + 2kg) * v'
15kg * m/s - 4kg * m/s = 5kg * v'
11kg * m/s = 5kg * v'
v' = 11kg * m/s / 5kg
v' = 11/5 m/s
Now, we can calculate the value of K2':
K2' = (1/2) * 5kg * (11/5 m/s)^2 = 24.2 J (joules)
Finally, let's calculate the kinetic energy lost:
Kinetic energy lost = (K1 + K2) - (K1' + K2')
= (37.5 J + 4 J) - (1.5 J + 24.2 J)
= 16.8 J
Therefore, the two-mass system loses 16.8 joules of kinetic energy during the collision.
amountLost= initial KE-final KE
= 1/2 *3*5^2+1/2 *2*2^2+1/2 *3*1^2=43Joules
check that.