Two objects, one of mass 3kg moving at 2m/s, the other of mass 5kg and speed of 2m/s move towards each other and collide in a head on collision. If the collision is perfectly inelastic, find the speed of the objects after the collision.
Since its perfectly inelastic the final velocity is negated since KE in the collision is maximum
M1V1-M2V2=(M1+M2)-Vc :Vc as final velocity
Substitute value
(3*2)-(5*2)=(5+3)(-Vc)
-4=-8Vc
Vc=0.5m/s
To find the speed of the objects after the collision, we can use the principle of conservation of momentum. In an inelastic collision, the total momentum before the collision is equal to the total momentum after the collision.
Let's assume that the 3kg object (object A) is moving to the right and the 5kg object (object B) is moving to the left. The initial momentum of object A is given by:
Initial momentum of A = mass of A × velocity of A
= 3 kg × 2 m/s
= 6 kg·m/s
The initial momentum of object B is given by:
Initial momentum of B = mass of B × velocity of B
= 5 kg × (-2 m/s)
= -10 kg·m/s (since B is moving in the opposite direction)
The total initial momentum is the sum of the initial momenta of the two objects:
Total initial momentum = Initial momentum of A + Initial momentum of B
= 6 kg·m/s - 10 kg·m/s
= -4 kg·m/s
Since momentum is conserved in the collision, the total final momentum after the collision is also -4 kg·m/s.
Let's assume that the objects stick together after the collision and move with a common final velocity, v. The final momentum after the collision is given by:
Total final momentum = (mass of A + mass of B) × velocity after the collision
= (3kg + 5kg) × v
= 8kg × v
Setting the total final momentum equal to -4 kg·m/s:
-4 kg·m/s = 8kg × v
Dividing both sides by 8kg:
v = -4 kg·m/s / 8kg
v = -0.5 m/s
Therefore, the objects will be moving with a final velocity of -0.5 m/s after the perfectly inelastic collision. Note that the negative sign indicates that they are moving in the opposite direction from their initial motion.
To solve this problem, we can use the law of conservation of momentum. According to this law, the total momentum before the collision is equal to the total momentum after the collision. This can be expressed as:
(m1 * v1) + (m2 * v2) = (m1 + m2) * vf
where:
m1 and m2 are the masses of the objects,
v1 and v2 are the initial velocities of the objects, and
vf is the final velocity of the objects after the collision.
Given:
m1 = 3 kg, v1 = 2 m/s (object 1)
m2 = 5 kg, v2 = -2 m/s (object 2, since it's moving in the opposite direction)
vf = ?
Using the equation above, we can plug in the values:
(3 kg * 2 m/s) + (5 kg * -2 m/s) = (3 kg + 5 kg) * vf
Simplifying further:
(6 kg * m/s) + (-10 kg * m/s) = (8 kg) * vf
-4 kg * m/s = 8 kg * vf
Dividing both sides of the equation by 8 kg:
vf = -4 kg * m/s / 8 kg
vf = -0.5 m/s
Therefore, the final velocity of the objects after the perfectly inelastic collision is -0.5 m/s.
call + x direction the initial direction of the first (3 kg) mass
initial momentum = 3*2 -5*2 =6-10 = -4
final momentum also -4, mass = 8
8 * v = -4
v = -1/2