A cart of mass M1 = 6.00 kg and initial speed = 4.00 m/s collides head on with a second cart of mass M2 = 4.00 kg at rest. Assuming that the collision is elastic, find the speed of M2 after the collision.
To find the speed of M2 after the collision, we can use the principle of conservation of momentum.
The conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. Mathematically, it can be written as:
m1v1(initial) + m2v2(initial) = m1v1(final) + m2v2(final)
Where:
m1 = mass of cart 1
v1(initial) = initial velocity of cart 1
m2 = mass of cart 2
v2(initial) = initial velocity of cart 2
v1(final) = final velocity of cart 1
v2(final) = final velocity of cart 2
In this case, cart 1 has an initial velocity of 4.00 m/s and cart 2 is at rest, so the equation becomes:
6.00 kg * 4.00 m/s + 4.00 kg * 0 m/s = 6.00 kg * v1(final) + 4.00 kg * v2(final)
Simplifying this equation gives:
24.00 kg m/s = 6.00 kg * v1(final) + 4.00 kg * v2(final)
Since the collision is assumed to be elastic, the kinetic energy is conserved. For an elastic collision, both momentum and kinetic energy are conserved.
Given that the initial kinetic energy is equal to the final kinetic energy, we can write the equation as:
(1/2) m1 * (v1(initial))^2 + (1/2) m2 * (v2(initial))^2 = (1/2) m1 * (v1(final))^2 + (1/2) m2 * (v2(final))^2
Substituting the values into the equation:
(1/2) * 6.00 kg * (4.00 m/s)^2 + (1/2) * 4.00 kg * (0 m/s)^2 = (1/2) * 6.00 kg * (v1(final))^2 + (1/2) * 4.00 kg * (v2(final))^2
Simplifying this equation gives:
48.00 kg m^2/s^2 = 3.00 kg * (v1(final))^2 + 0.00 kg * (v2(final))^2
Since the initial velocity of cart 2 is zero, the term (1/2) * 4.00 kg * (v2(final))^2 becomes zero.
Now, the equation becomes:
48.00 kg m^2/s^2 = 3.00 kg * (v1(final))^2
Rearranging the equation to solve for (v1(final))^2:
(v1(final))^2 = 48.00 kg m^2/s^2 / 3.00 kg
Simplifying this equation gives:
(v1(final))^2 = 16.00 m^2/s^2
Taking the square root of both sides gives the final velocity of cart 1:
v1(final) = √16.00 m^2/s^2
v1(final) = 4.00 m/s
Since the collision is elastic and we now have the final velocity of cart 1, we can find the final velocity of cart 2 by substituting this value into the momentum conservation equation:
6.00 kg * 4.00 m/s + 4.00 kg * 0 m/s = 6.00 kg * 4.00 m/s + 4.00 kg * v2(final)
Rearranging the equation to solve for v2(final):
4.00 kg * v2(final) = 6.00 kg * 4.00 m/s - 6.00 kg * 4.00 m/s
Simplifying this equation gives:
4.00 kg * v2(final) = 48.00 kg m/s - 24.00 kg m/s
4.00 kg * v2(final) = 24.00 kg m/s
Dividing both sides by 4.00 kg gives:
v2(final) = 24.00 kg m/s / 4.00 kg
v2(final) = 6.00 m/s
Therefore, the speed of M2 after the collision is 6.00 m/s.