Consider a frictionless track as shown in the figure below. A block of mass m1 = 5.95 kg is released from A. It makes a head on elastic collision at B with a block of mass m2 = 10.0 kg that is initially at rest. Calculate the maximum height to which m1 rises after the collision. I need the formula so i can work it out..thanks
To calculate the maximum height to which m1 rises after the collision, you can use the principles of conservation of mechanical energy. Here's how you can derive the formula:
1. Assume that the initial velocity of m1 is u1 and the final velocity after the collision is v1. Since the collision is elastic, the relative velocity of m1 and m2 after the collision is the negative of their relative velocity before the collision.
2. Apply conservation of linear momentum in the horizontal direction: m1 * u1 = m1 * v1 - m2 * (-v1). Simplifying this equation gives us m1 * u1 = (m1 + m2) * v1.
3. Apply conservation of mechanical energy for the block m1. The initial mechanical energy is purely kinetic energy and the final mechanical energy is purely potential energy. Therefore, we can write: (1/2) * m1 * u1^2 = m1 * g * h_max, where g is the acceleration due to gravity and h_max is the maximum height.
4. Rearrange this equation to solve for h_max: h_max = (1/2) * u1^2 / g.
Given that m1 = 5.95 kg, m2 = 10.0 kg, and the acceleration due to gravity g = 9.8 m/s^2, you can substitute these values into the equation along with the initial velocity u1 to find the maximum height h_max.
I suggest plugging in the actual values into the formula to calculate the maximum height.