Two bodies A and B of mass 4kg and 2kg moves toward each other with velocity 3mls and 2mls and collides, if the collision is perfectly inelastic, find the velocity of the two bodies.after collision find the total k.e of the system before and after collision,hence calculate the loss in k.e
you need to specify the velocities, not just the speeds.
to be submitted
To find the velocity of the two bodies after the collision, we can use the principle of conservation of momentum. According to this principle, the total momentum of an isolated system remains constant before and after a collision.
Let's denote the velocity of body A after the collision as 'vA' and the velocity of body B after the collision as 'vB'.
Step 1: Find the initial momentum before the collision:
Momentum of A before collision = Mass of A * Velocity of A
= 4 kg * 3 m/s
= 12 kg·m/s
Momentum of B before collision = Mass of B * Velocity of B
= 2 kg * (-2 m/s) [Note: Negative sign indicates the opposite direction]
= -4 kg·m/s
The total initial momentum before the collision is the sum of the individual momenta:
Initial momentum = (12 kg·m/s) + (-4 kg·m/s)
= 8 kg·m/s
Step 2: Apply the principle of conservation of momentum:
Total momentum after the collision = Initial momentum
Momentum of A after collision = Mass of A * Velocity of A after collision
= 4 kg * vA
Momentum of B after collision = Mass of B * Velocity of B after collision
= 2 kg * vB
Now, according to the principle of conservation of momentum:
(4 kg * vA) + (2 kg * vB) = 8 kg·m/s
Step 3: Apply the condition of perfectly inelastic collision:
In a perfectly inelastic collision, the bodies stick together after the collision. This means that their velocities become the same.
vA = vB (let's call it v)
By substituting this condition into the conservation of momentum equation, we can solve for v:
(4 kg * v) + (2 kg * v) = 8 kg·m/s
6 kg·v = 8 kg·m/s
v = 8 kg·m/s / 6 kg
v ≈ 1.33 m/s
Therefore, the velocity of both bodies A and B after the collision is approximately 1.33 m/s.
To find the total kinetic energy (KE) of the system before and after the collision:
Step 4: Calculate the initial kinetic energy:
Kinetic energy of A before collision = (1/2) * Mass of A * (Velocity of A)^2
= (1/2) * 4 kg * (3 m/s)^2
= 18 J
Kinetic energy of B before collision = (1/2) * Mass of B * (Velocity of B)^2
= (1/2) * 2 kg * (-2 m/s)^2
= 4 J [Note: Negative sign does not affect KE calculation]
Total kinetic energy before collision = KE of A + KE of B
= 18 J + 4 J
= 22 J
Step 5: Calculate the kinetic energy after the collision:
Since the bodies stick together after the collision, their velocities become the same (v ≈ 1.33 m/s).
Kinetic energy after collision = (1/2) * (Mass of A + Mass of B) * (Velocity after collision)^2
= (1/2) * (4 kg + 2 kg) * (1.33 m/s)^2
= 6 J
Finally, we can calculate the loss in kinetic energy:
Loss in kinetic energy = Total initial kinetic energy - Total kinetic energy after collision
= 22 J - 6 J
= 16 J
So, the loss in kinetic energy during the perfectly inelastic collision is 16 J.
To find the velocity of the two bodies after the perfectly inelastic collision, we can use the conservation of momentum principle. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
The total momentum before the collision is given by the sum of the individual momenta of the two bodies:
Momentum of body A (pA) = mass of A (mA) * velocity of A (vA)
= 4 kg * 3 m/s
= 12 kg m/s
Momentum of body B (pB) = mass of B (mB) * velocity of B (vB)
= 2 kg * (-2) m/s (since B is moving toward A)
= -4 kg m/s
Total momentum before collision:
p_initial = pA + pB
= 12 kg m/s - 4 kg m/s
= 8 kg m/s
Since the collision is perfectly inelastic, the two bodies stick together after the collision and move as one body. Let's call the mass of the combined bodies as m_final, and the velocity of the combined bodies after the collision as v_final.
Total momentum after collision:
p_final = m_final * v_final
Since momentum is conserved, p_initial = p_final.
So, we can equate the two expressions:
8 kg m/s = (4 kg + 2 kg) * v_final
8 kg m/s = 6 kg * v_final
Now we can solve for v_final:
v_final = 8 kg m/s / 6 kg
v_final = 4/3 m/s
Therefore, after the perfectly inelastic collision, the velocity of the two bodies is 4/3 m/s.
Now let's calculate the total kinetic energy (KE) of the system before and after the collision:
Total KE before collision:
KE_initial = (1/2) * mA * vA^2 + (1/2) * mB * vB^2
= (1/2) * 4 kg * (3 m/s)^2 + (1/2) * 2 kg * (2 m/s)^2
= 18 J + 4 J
= 22 J
Total KE after collision:
KE_final = (1/2) * m_final * v_final^2
= (1/2) * 6 kg * (4/3 m/s)^2
= (1/2) * 6 kg * (16/9) m^2/s^2
= 32/3 J
Finally, we can calculate the loss in kinetic energy:
Loss in KE = KE_initial - KE_final
= 22 J - 32/3 J
= 66/3 J - 32/3 J
= 34/3 J
Therefore, the loss in kinetic energy during the collision is 34/3 J.