A cart of mass M1 = 4.00 kg and initial speed = 4.00 m/s collides head on with a second cart of mass M2 = 2.00 kg at rest. Assuming that the collision is elastic, find the speed of M2 after the collision.
i keep getting 2.3 m/s
what am i doing wrong?
Show your work and we'll tell you. Did you assume that both total momentum and kinetic energy were conserved?
The center of mass moves at velocity 2/3* 4.00 = 8/3 m/s. In that "CM" coordinate system, M2 approaches with velocity -8/3 m/s and leaves with velocity +8/3 m/s. In earth-fixed coordinates, its speed is 8/3 + 8/3 = 16/3 = 5.33 m/s.
The speed of M1 after collision is -(4 - 8/3) + 8/3 = 4/3 = 1.33 m/s.
You may verify that these numbers are correct by plugging them into the momentum and KE conservation equations.
okay I think I get it now, I was going at it in a different way.
thanks so much.
To solve this problem, we can apply the principle of conservation of momentum and kinetic energy. Let's break down the steps:
1. Write down the initial and final momentum equations separately for M1 and M2:
Initial momentum of M1 = M1 * v1 (where v1 is the initial velocity of M1)
Initial momentum of M2 = 0 (since M2 is at rest initially)
Final momentum of M1 = M1 * v1' (where v1' is the final velocity of M1)
Final momentum of M2 = M2 * v2 (where v2 is the final velocity of M2)
2. Write down the conservation of momentum equation:
Initial momentum of M1 + Initial momentum of M2 = Final momentum of M1 + Final momentum of M2
M1 * v1 + 0 = M1 * v1' + M2 * v2
3. Rearrange the equation to solve for v2:
v2 = (M1 * v1 - M1 * v1') / M2
4. Since the collision is elastic, kinetic energy is conserved:
Initial kinetic energy = Final kinetic energy
0.5 * M1 * v1^2 = 0.5 * M1 * v1'^2 + 0.5 * M2 * v2^2
5. Substitute the value of v2 from step 3 into the equation above and solve for v1':
0.5 * M1 * v1^2 = 0.5 * M1 * v1'^2 + 0.5 * M2 * ((M1 * v1 - M1 * v1') / M2)^2
Simplify the equation and solve for v1':
v1'^2 - 2 * v1 * v1' + v1^2 = 0
6. Solve the quadratic equation from step 5 to find v1':
v1' = (v1 + sqrt(v1^2 - v1^2)) / 2
= (v1 + 0) / 2
= v1 / 2
Therefore, the correct answer for the final velocity of M2 (v2) is (v1 / 2), which would be (4.00 m/s / 2) = 2.00 m/s.
Your answer of 2.3 m/s is incorrect. The correct final velocity for M2 is 2.00 m/s.
To find the speed of M2 after the collision, we can use 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 in an isolated system.
The momentum of an object is defined as the product of its mass and velocity. So, the initial momentum before the collision can be calculated as follows:
Initial momentum of M1 = M1 * v1 (where v1 is the initial speed of M1)
Initial momentum of M2 = M2 * 0 (since M2 is initially at rest)
Since there are no other external forces acting on the system, the total momentum before the collision is simply the sum of the individual momenta:
Total initial momentum = (M1 * v1) + (M2 * 0)
= M1 * v1
After the collision, let's assume M1 moves with a speed v1' and M2 moves with a speed v2. Applying the conservation of momentum, the total momentum after the collision is given by:
Total final momentum = (M1 * v1') + (M2 * v2)
Since the collision is elastic, kinetic energy is conserved as well. The kinetic energy of an object is given by half the product of its mass and the square of its velocity. So we can write the equation for the conservation of kinetic energy as:
Initial kinetic energy = Final kinetic energy
Initial kinetic energy = (1/2) * M1 * v1^2 + 0 (since M2 is initially at rest)
Final kinetic energy = (1/2) * M1 * (v1')^2 + (1/2) * M2 * v2^2
Now, using the principles of momentum and kinetic energy conservation, we can solve for the unknowns. Plugging in the given values M1 = 4.00 kg, v1 = 4.00 m/s, M2 = 2.00 kg:
M1 * v1 = M1 * v1' + M2 * v2 (momentum conservation)
(1/2) * M1 * v1^2 = (1/2) * M1 * (v1')^2 + (1/2) * M2 * v2^2 (kinetic energy conservation)
Substituting the known values:
4.00 kg * 4.00 m/s = 4.00 kg * v1' + 2.00 kg * v2
8.00 kg*m/s = 4.00 kg * (v1')^2 + 2.00 kg * v2^2
It appears that your result of 2.3 m/s may be incorrect. Solving the equation above, we get:
16.00 kg*m/s = 4.00 kg * (v1')^2 + 2.00 kg * v2^2
To find v2, we need to solve for v1' first. Rearranging the equation, we have:
4.00 kg * (v1')^2 = 16.00 kg*m/s - 2.00 kg * v2^2
(v1')^2 = 4.00 m^2/s^2 - 0.50 v2^2
v1' = √(4.00 m^2/s^2 - 0.50 v2^2)
Now, substitute v1' into the equation for momentum conservation:
8.00 kg*m/s = 4.00 kg * √(4.00 m^2/s^2 - 0.50 v2^2) + 2.00 kg * v2
To solve this equation, you can square both sides and isolate v2 on one side. Then, square again to find the square root of v2^2 and get the final value for v2.
Make sure to double-check your calculations and confirm if your approach and equations align with the problem statement.