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 ? (3 marks)
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Given A ball 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 m/s^2 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 ?
- So given initial velocity u = 0.600
- So the ball comes to a stop, v = 0 m/s
- So the distance is 2.00 m
- So x – xo = 2.00 x 10^-3 m
- So we have the equation of motion v^2 = u^2 + 2as
- Or a = v^2 – u^2 / 2s
- deceleration a = v^2 – u^2 / 2(x – xo)
- = 0 ^2 – (0.600)^2 / 2 (2.00 x 10^-3)
- = - 90 m / s^2
- Now we need to find deceleration in multiples of g, so we have
- = mod a/g
- = 90 / 9.81
- = 9.18
- Now the stopping time will be
- x – xo = ½ (u + v)t
- Or t = 2 (x – xo) / u + v
- = 2(2.00 x 10^-3) / 0.600 + 0
- = 6.67 x 10^-3 secs
- Now we have
- x – xo = 4.50 mm
- = 4.50 x 10^-3 m
- a = v^2 – u^2 / 2 (x – x0)
- = 0 – 0.600 / 2(4.50 x 10^-3 m)
- a = - 40 m/s^2
- So in multiples of g we get a/g = 40 / 9.8
- = 4.08
Reference link will be
https://brainly.in/question/17064578
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