1- A 5.0 kg block moving east with speed of 6 m/s collides head on with another 8.0 kg block moving west with speed of 4 m/s. After the collision the 5.0 kg block was moving west with speed of 5 m/s while the 8.0 kg block was moving east.
a) What is the speed of the 8.0 kg block after the collision?
b) Determine whether the collision was elastic or inelastic.
(Show complete work)
6 m/s
5 m/s
Before collision
After collision
4 m/s
v =?
5.0 Kg
8.0 Kg
5.0 Kg
8.0 Kg
a) To find the speed of the 8.0 kg block after the collision, we can use the principle of conservation of momentum.
Before the collision, the total momentum of the system is given by:
Initial momentum = (mass of 5.0 kg block * velocity of 5.0 kg block) + (mass of 8.0 kg block * velocity of 8.0 kg block)
= (5.0 kg * 6 m/s) + (8.0 kg * (-4 m/s))
= 30 kg m/s - 32 kg m/s
= -2 kg m/s (Note: The negative sign indicates the opposite direction of motion)
After the collision, the total momentum of the system is also conserved:
Final momentum = (mass of 5.0 kg block * velocity of 5.0 kg block) + (mass of 8.0 kg block * velocity of 8.0 kg block)
= (5.0 kg * (-5 m/s)) + (8.0 kg * v) (Note: The velocity of the 5.0 kg block has changed to -5 m/s)
= -25 kg m/s + 8v
Since momentum is conserved, we can equate the initial and final momenta:
-2 kg m/s = -25 kg m/s + 8v
Simplifying the equation:
8v = -2 kg m/s + 25 kg m/s
8v = 23 kg m/s
v = 23 kg m/s / 8
v ≈ 2.875 m/s
Therefore, the speed of the 8.0 kg block after the collision is approximately 2.875 m/s.
b) To determine whether the collision was elastic or inelastic, we need to calculate the kinetic energy before and after the collision.
The initial kinetic energy is given by:
Initial KE = (1/2) * (mass of 5.0 kg block) * (velocity of 5.0 kg block)^2 + (1/2) * (mass of 8.0 kg block) * (velocity of 8.0 kg block)^2
= (1/2) * 5.0 kg * (6 m/s)^2 + (1/2) * 8.0 kg * (4 m/s)^2
= 90 J + 64 J
= 154 J
The final kinetic energy is given by:
Final KE = (1/2) * (mass of 5.0 kg block) * (velocity of 5.0 kg block)^2 + (1/2) * (mass of 8.0 kg block) * (velocity of 8.0 kg block)^2
= (1/2) * 5.0 kg * (5 m/s)^2 + (1/2) * 8.0 kg * (2.875 m/s)^2
= 62.5 J + 82 J
= 144.5 J
Since the initial kinetic energy (154 J) is not equal to the final kinetic energy (144.5 J), the collision is inelastic.
To solve this problem, we can use the law of conservation of momentum:
Before the collision, we have:
Momentum of the 5.0 kg block (moving east): P1 = (mass 1) * (velocity 1) = (5.0 kg) * (6 m/s) = 30 kg*m/s
Momentum of the 8.0 kg block (moving west): P2 = (mass 2) * (velocity 2) = (8.0 kg) * (-4 m/s) = -32 kg*m/s (negative sign because it is moving in the opposite direction)
Total momentum before the collision: P_total = P1 + P2 = 30 kg*m/s - 32 kg*m/s = -2 kg*m/s
After the collision, we have:
Momentum of the 5.0 kg block (moving west): P1' = (mass 1) * (velocity 1') = (5.0 kg) * (-5 m/s) = -25 kg*m/s
Momentum of the 8.0 kg block (moving east): P2' = (mass 2) * (velocity 2') = (8.0 kg) * (v m/s) (we need to find v in part a)
Total momentum after the collision: P_total' = P1' + P2' = -25 kg*m/s + (8.0 kg) * (v m/s)
According to the law of conservation of momentum, the total momentum before the collision should be equal to the total momentum after the collision:
P_total = P_total'
-2 kg*m/s = -25 kg*m/s + (8.0 kg) * (v m/s)
Now we can solve for v:
-2 kg*m/s + 25 kg*m/s = (8.0 kg) * (v m/s)
23 kg*m/s = (8.0 kg) * (v m/s)
v = 23 kg*m/s / 8.0 kg
v = 2.875 m/s
So, the speed of the 8.0 kg block after the collision is 2.875 m/s.
To determine whether the collision was elastic or inelastic, we need to compare the total kinetic energy before and after the collision. If the kinetic energy is conserved, the collision is elastic. If some of the kinetic energy is lost, the collision is inelastic.
Before the collision, the total kinetic energy is given by:
KE_before = (1/2) * (mass 1) * (velocity 1)^2 + (1/2) * (mass 2) * (velocity 2)^2
KE_before = (1/2) * (5.0 kg) * (6 m/s)^2 + (1/2) * (8.0 kg) * (4 m/s)^2
KE_before = 90 J + 64 J
KE_before = 154 J
After the collision, the total kinetic energy is given by:
KE_after = (1/2) * (mass 1) * (velocity 1')^2 + (1/2) * (mass 2) * (velocity 2')^2
KE_after = (1/2) * (5.0 kg) * (-5 m/s)^2 + (1/2) * (8.0 kg) * (2.875 m/s)^2
KE_after = 62.5 J + 82 J
KE_after = 144.5 J
Since KE_before (154 J) is not equal to KE_after (144.5 J), we can conclude that some kinetic energy was lost during the collision.
Therefore, the collision was inelastic.