Question

Two blocks of masses m1 = 2.40 kg and m2 = 4.80 kg are each released from rest at a height of h = 5.10 m on a frictionless track, as shown in the figure below, and undergo an elastic head-on collision. (Let the positive direction point to the right. Indicate the direction with the sign of your answer.) Two blocks are on a curved ramp similar in shape to a half-pipe. There is a flat horizontal surface with opposite ends curving upwards. A block of mass m1 is on the left curved end at a height h above the lowest point of the track. A block of mass m2 is on the right curved end also at a height h above the lowest point of the track. (a) Determine the velocity of each block just before the collision. v1i = m/s v2i = m/s (b) Determine the velocity of each block immediately after the collision. v1f = m/s v2f = m/s. Determine the maximum heights to which m1 and m2 rise after the collision. y1f = m y2f = m

Answer 1

To determine the velocities of the blocks before and after the collision, we can use the principles of conservation of mechanical energy and linear momentum. By setting the initial potential energy equal to the final kinetic energy, we can find the velocities before the collision. Then, by using the conservation of linear momentum, we can solve for the velocities after the collision.

Step-by-step explanation:

Given the scenario in which two blocks of masses m1 = 2.40 kg and m2 = 4.80 kg are released from rest at a height of h = 5.10 m on a frictionless track, we can determine the velocities of the blocks before and after the collision.

To calculate the velocities before the collision, we can use the principle of conservation of mechanical energy. Since the blocks are released from rest at a height h, their initial potential energy is given by mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. Setting the initial potential energy equal to the final kinetic energy, we have:

m1gh = (1/2)m1v1^2 + (1/2)m2v2^2

Using the principle of conservation of linear momentum, we can determine the velocities of the blocks after the collision. Since the collision is elastic, the total momentum before the collision is equal to the total momentum after the collision:

m1v1i + m2v2i = m1v1f + m2v2f

Using these equations, we can solve for the velocities before and after the collision.

Answer 2

To determine the velocities of the blocks before and after the collision, conservation of energy is used to calculate the initial velocities, and conservation of momentum and kinetic energy is used to find the final velocities.

Step-by-step explanation:

To find the velocity of each block just before the collision, we use conservation of energy. Since the blocks start from rest, the potential energy (PE) at height h will be converted to kinetic energy (KE) just before the collision. For block m1: PE = KE, giving m1gh = (1/2)m1v1i2, so v1i = sqrt(2gh). Plugging in g = 9.8 m/s2 and h = 5.10 m, we find v1i = sqrt(2*9.8*5.10) = sqrt(99.96) ≈ 9.998 m/s. Similarly, for block m2, we calculate v2i = -9.998 m/s, negative because it's going to the left.

Now, we use the conservation of momentum and kinetic energy to determine the velocities after the elastic collision. For block m1 and m2 after the collision (v1f and v2f respectively), the momentum before the collision equals the momentum after the collision, and the total kinetic energy before collision equals the total kinetic energy after the collision:

1. m1v1i + m2v2i = m1v1f + m2v2f (Momentum conservation)
2. (1/2)m1v1i2 + (1/2)m2v2i2 = (1/2)m1v1f2 + (1/2)m2v2f2 (Kinetic energy conservation)

By solving these equations, we can find v1f and v2f. After the collision, each block will rise to a height where all its kinetic energy is again converted to potential energy, so at maximum height, KE = PE, and we solve for the new heights y1f and y2f using the respective velocities v1f and v2f.