Question

A block of mass 0.20kg is initially on a curved frictionless ramp at height h0 = 75.0 cm. It is given a push so that at this height it has speed 2.00 m/s. At the bottom of the curved ramp, there is a large block (mass 1.00 kg) which is initially at rest. The two blocks collide elastically - the small block bounces backward, and the larger block moves forward with a speed 1.441 m/s. What is the maximum height acheived by the small block as it heads back up the ramp?

Answer 1

To find the maximum height achieved by the small block as it heads back up the ramp, we can use the principle of conservation of energy. At the highest point of the block's trajectory, it has converted its kinetic energy into potential energy. The maximum height can be calculated using the equation h1 = (v^2) / (2g), where v is the initial velocity and g is the acceleration due to gravity.

Step-by-step explanation:

To solve this problem, we can use the principle of conservation of energy. At the highest point of the block's trajectory, it has converted its kinetic energy into potential energy. The potential energy at this point is equal to the gravitational potential energy at the initial height. We can use this relationship to find the maximum height achieved by the small block.

Let's denote the maximum height as h1. The initial kinetic energy of the block is given as 1/2 mv2, where m is the mass of the block and v is the initial velocity. At the maximum height, the block has zero kinetic energy, so all of the initial kinetic energy has been converted into potential energy:

1/2 mv2 = mgh1

Using this equation, we can solve for h1:

h1 = (v2) / (2g)

Plugging in the values given in the problem, we have v = 2.00 m/s and g is the acceleration due to gravity (9.8 m/s2). Solving for h1:

h1 = (2.002) / (2 * 9.8) = 0.204 m

Therefore, the maximum height achieved by the small block as it heads back up the ramp is 0.204 m.

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