Question

A 0.500 kg ball is dropped from rest at a point 1.20 m above the floor. The ball rebounds straight upward to a height of 0.700 ma) What is the velocity one instant before the collision?b) what is the velocity one instant after the collision with the floor?c) what is the magnitude and direction of the impulse of the net force applied to the ball during the collision with the floor?

Answer 1

To solve this problem we will apply the concepts related to energy conservation. With this we will find the speed before the impact. Through the kinematic equations of linear motion we will find the velocity after the impact.

Since the momentum is given as the product between mass and velocity difference, we will proceed with the velocities found to calculate it.

Part A) Conservation of the energy

$mgh = \frac{1}{2} mv^2$

$v_1 = \sqrt{2gh}$

$v_1 = \sqrt{2(9.8)(1.20)}$

$v_1 = 4.84m/s$

$\therefore v_1 = -4.84/s \rightarrow \text{Negative direction downward}$

Part B)  Kinematic equation of linear motion,

$v_2^2 = u_0^2 +2a\Delta y$

Here

v= 0 Because at 1.5m reaches highest point, so v=0

$0 = u_2^2 +2(-9.8)(0.7)$

$u_2 = 3.7m/s$

Therefore the velocity after the collision with the floor is 3.7m/s

PART C) Total change of impulse is given as,

$J = P_2 -P_1$

$J = mU_2-mV_1$

$J = m(U_2-V_1)$

$J = (0.5)(3.7-(-4.84))$

$J = 4.27kg \cdot m/s \rightarrow \text{Upward because is positive}$