Question

Two blocks are free to slide along the frictionless wooden track shown below. the block of mass m1 = 4.97 kg is released from the position shown, at height h = 5.00 m above the flat part of the track. protruding from its front end is the north pole of a strong magnet, which repels the north pole of an identical magnet embedded in the back end of the block of mass m2 = 10.8 kg, initially at rest. the two blocks never touch. calculate the maximum height to which m1 rises after the elastic collision. m two masses are shown on a frictionless wooden track. the left part of the track curves downward from left to right, starting from an almost-vertical slope and then decreasing in slope magnitude until it is horizontal. the right part of the track is horizontal and has a longer surface length than that of the first part of the track. mass m1 is on the curved part of the track a vertical distance h from the lowest point on the track. mass m2 is on the second part of the track just where the track becomes horizontal.

Answer 1

The maximum height h_2 to which m_1 rises after the elastic collision is given by h_2 = 1} / 2 m_2 / m_1 + 1 h.

Step-by-step explanation:

To calculate the maximum height to which mass m_1 rises after the elastic collision, we can use the conservation of mechanical energy.

The initial potential energy of m_1 is converted into kinetic energy, and then, after the collision, it is converted back into potential energy. The equation for conservation of energy is:

m_1gh = {1} / {2} m_1 v_1^2 + {1} / {2} m_2 v_2^2

where:

- m_1 is the mass of block 1,

- m_2 is the mass of block 2,

- h is the initial height,

- g is the acceleration due to gravity,

- v_1 is the velocity of block 1 after the collision,

- v_2 is the velocity of block 2 after the collision.

Since the collision is elastic, the relative velocity of approach before the collision is equal to the relative velocity of separation after the collision. Therefore, v_2 - v_1 = u_2 - u_1, where u_1 and u_2 are the initial velocities of m_1 and m_2 respectively.

u_1 is the velocity acquired by m_1 when it reaches the bottom of the slope, and u_2 is the velocity of m_2 initially at rest.

The energy at the bottom of the slope is entirely potential energy, so (u_1 = {2gh}^1/2). Since (m_2) is initially at rest, u_2 = 0.

Now, we can set up the equation using the conservation of energy and the relative velocity equation:

m_1gh = {1} / {2} m_1 v_1^2 + {1} / {2} m_2 v_2^2

m_1gh = {1} / {2} m_1 v_1^2 + {1} / {2} m_2 (v_1)^2

2gh = v_1^2 {m_1} / {m_1} + {m_2} / {m_1}

Now, solve for v_1:

v_1 = {{2gh} / {m_1} / {m_1} + {m_2} / {m_1}}^1/2

Once you find v_1, you can use it to calculate the maximum height h_2 reached by m_1 after the collision using the conservation of energy:

m_1gh_2 = {1} / {2} m_1 (v_1)^2

h_2 = {v_1^2} / {2g}

Plug in the values to find h_2.