Question

An iron sphere of mass 5 kg is dropped from a height of 20 m if the acceleration of the sphere is 9.8 metre per second square calculate the momentum transferred to the ground by the ball

Answer 1

Mass of the sphere = 5 kg

Height from with the sphere is dropped = 20 m

Acceleration = 20 $\(\frac{m}{s^2}\)$

Initial velocity = 0 $\(\frac{m}{s}\)$

Let the final velocity just before it hits the ground be v.

Now, using the third equation of motion:

$v_f^2 = v_i^2 + 2 \times a \times H$

$\(v_f^2 = 0^2 + 2 \times 9.8 \times 20\)$

$\(v_f^2 = 392\)$

$\(v_f = 19.79 \frac{m}{s}\)$

Momentum of the ball just before it hits the ground = $\( mass \times v_f \)$

Momentum transferred to the ground = $\( mass \times v_f \)$

= 5 × 19.79

= $\(98.95\) \(kg\frac{m}{s}\)$

Momentum transferred to the ground is $\(98.95\) \(kg\frac{m}{s}\)$