Question

A student sends vertically to the floor, from the height H1 = 1,35m, a ball with the mass m = 400 g. The initial speed of the ball is V0 = 3M / s.

. Immediately after collision with the floor, the ball speed is facing vertically and

It is a fraction K of the ball's speed immediately before hitting the floor. The ball goes back up up to the maximum height H2 = 1.25M

. The air friction forces are negligible. Potential gravitational energy is considered null at the floor level, and the dimensions of the ball are neglected. Determine:

a. Total mechanical energy of the ball at the time of launch;

b. the ball pulse value immediately after collision with the floor;

Answer 1

Solution.

$H_1=1.35m;$

$m=400g=0.4kg;$

$v_0=3m/s;$

$H_2=1.25m;$

$a) W=mgH_1+\dfrac{mv_0^2}{2};$

$W=0.4kg\sdot9.8N/kg\sdot1.35m+\dfrac{0.4kg\sdot(3m/s)^2}{2}=7.092J;$

$b) mgH_2=\dfrac{mv^2}{2}\implies v=\sqrt{2gH_2};$

$v=\sqrt{2\sdot9.8N/kg\sdot1.25m}=4.95m/s;$

$p=mv;$

$p=0.4kg\sdot4.95m/s=1.98kg\sdot m/s;$

Answer:$a) W=7.092J;$

$b) p=1.98kg\sdot m/s$ .