Question

B- Aball comes to a stop from an initial velocity of 0.600 m/s in a distance of only 2.00
mm.
(a) Find the acceleration in
2 m/s and in multiples of  2  g g  9.80 m/s . (3 marks)
(b) Calculate the stopping time. (3 marks)
(c) If the stopping distance is 4.50 mm. What is the deceleration, expressed in multiples of g ?

Answer 1

Given info : A ball comes to stop from an initial velocity of 0.6 m/s in a distance of only 2 mm.

To find :

1. find acceleration in m/s² and in multiple of g
2. calculate the stopping time
3. if the stopping distance is 4.5 mm what is deceleration, expressed in multiple of g

solution : 1. using formula, v² = u² + 2as

here v = 0, u = 0.6 m/s , s = 2 mm = 0.002 m

now, 0² = (0.6)² + 2 × a × 0.002

⇒a = -0.36/0.004 = -90 m/s²

now acceleration in g = -(90/9.8) g ≈ -9.2g

2. using formula, v = u + at

⇒0 = 0.6 + (-90)t

⇒t = 0.6/90 = 0.0067 sec

3. stopping distance, s = 4.5 mm = 0.0045 m

using formula, v² = u² + 2as

⇒0² = (0.6)² + 2 × a × 0.0045

⇒a = -0.36/(0.009) = -40 m/s²

so deceleration is 40 m/s²

now deceleration in g = 40 × g/9.8 = 4.08g