Question

A basketball has a mass of 567 g. Heading straight downward, in the direction, it hits the floor with a speed of 2 m/s and rebounds straight up with nearly the same speed. What was the momentum change ?

Answer 1

Answer:

$\Delta p=2.27\frac{kg\cdot m}{s}$

Explanation:

The momentum change is defined as:

$\Delta p=p_f-p_i\\\Delta p=mv_f-mv_i\\\Delta p=m(v_f-v_i)(1)$

Taking the downward motion as negative and the upward motion as positive, we have:

$v_f=2\frac{m}{s}(2)\\v_i=-2\frac{m}{s}(3)$

Replacing (2) and (3) in (1):

$\Delta p=567*10^{-3}kg(2\frac{m}{s}-(-2\frac{m}{s}))\\\Delta p=2.27\frac{kg\cdot m}{s}$

Answer 2

The change in momentum is directly proportional to the change in velocity. The change in momentum of the given basketball is 2.27 kgm/s.

Conservation of momentum:

$\Delta p = m (v_f - v_i)$

Where,

$\Delta p$ - change in momentum

$m$ - mass  = 567 g

$v_f$ - final velocity= -2 m/s

$v_i$ - initial velocity= 2 m/s

Put the values in the formula,

$\Delta p = 567 (-2 - 2)\\\\ \Delta p = 2.27\rm\ kgm/s.$

Therefore, the change in momentum of the given basketball is 2.27 kgm/s.

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