a 5 kg block with a speed of 3 m/s collides with a 10 kg block that has a speed of 2 m/s in the same direction. After the collision, the 10 kg block is observed to be traveling in the original direction with a speed of 2.5 m/s.
a) what is the speed of the 5 kg block immediately after the collision?
b) By how much does the total kinetic energy of the system of the 2 blocks change because of the collision?
c)suppose, instead, that the 10 kg block ends up with a speed of 4 m/s. what then is the change in total KE?
a) To find the speed of the 5 kg block immediately after the collision, we need to apply the principle of conservation of momentum. According to this principle, the total momentum before the collision should be equal to the total momentum after the collision.
The momentum of an object is given by the product of its mass and velocity, which can be represented as:
Momentum = mass * velocity
Before the collision, the momentum of the 5 kg block is:
Momentum1 = mass1 * velocity1 = 5 kg * 3 m/s = 15 kg·m/s
Before the collision, the momentum of the 10 kg block is:
Momentum2 = mass2 * velocity2 = 10 kg * 2 m/s = 20 kg·m/s
Using the principle of conservation of momentum, we have:
Total momentum before collision = Total momentum after collision
15 kg·m/s + 20 kg·m/s = (5 kg * v1) + (10 kg * 2.5 m/s)
Simplifying the equation, we get:
35 kg·m/s = 5 kg * v1 + 25 kg·m/s
Rearranging the equation, we find:
5 kg * v1 = 35 kg·m/s - 25 kg·m/s
5 kg * v1 = 10 kg·m/s
Dividing both sides of the equation by 5 kg, we get:
v1 = 10 kg·m/s / 5 kg
v1 = 2 m/s
Therefore, the speed of the 5 kg block immediately after the collision is 2 m/s.
b) To calculate the change in total kinetic energy of the system, we need to compare the initial kinetic energy (before the collision) with the final kinetic energy (after the collision).
The kinetic energy of an object is given by the formula:
Kinetic Energy = (1/2) * mass * velocity^2
Before the collision, the total kinetic energy of the system is:
Initial KE = (1/2) * (5 kg) * (3 m/s)^2 + (1/2) * (10 kg) * (2 m/s)^2
Initial KE = 22.5 J + 20 J
Initial KE = 42.5 J
After the collision, the total kinetic energy of the system is:
Final KE = (1/2) * (5 kg) * (2 m/s)^2 + (1/2) * (10 kg) * (2.5 m/s)^2
Final KE = 10 J + 31.25 J
Final KE = 41.25 J
The change in total kinetic energy is given by:
Change in KE = Final KE - Initial KE
Change in KE = 41.25 J - 42.5 J
Change in KE = -1.25 J
Therefore, the total kinetic energy of the system decreases by 1.25 Joules because of the collision.
c) If the 10 kg block ends up with a speed of 4 m/s instead, we can again find the change in total kinetic energy using the same steps as above.
Using the final speed of 10 kg block as 4 m/s, we can calculate the change in total kinetic energy:
Final KE = (1/2) * (5 kg) * (2 m/s)^2 + (1/2) * (10 kg) * (4 m/s)^2
Final KE = 10 J + 80 J
Final KE = 90 J
Change in KE = Final KE - Initial KE
Change in KE = 90 J - 42.5 J
Change in KE = 47.5 J
Therefore, if the 10 kg block ends up with a speed of 4 m/s, the change in total kinetic energy of the system would be 47.5 Joules.