Question

Problem 1. You hit a bowling ball with a mallet, starting from rest. Every time you hit it, its velocity increases by exactly 10 cm/s. You start timing (t 0 seconds) at the instant you hit the ball the first time, and you hit it once every second. Two simplifying approximations: (1) Suppose each hit is very short, so the speed changes in a short amount of time and (2) suppose the ball doesn't slow down between hits a. Att - 5 seconds, how far has the ball moved from its starting position? b. Now you ease up how hard you hit the ball, so that it speeds up 5 cm/s with every hit. But now you hit it once every half second instead. Start the clock again at the instant you hit the ball the first time, starting from rest, and again find how far the ball has moved at t 5 seconds. (Hint: Can you find a pattern rather than just adding up each small segment?) c. Plot the displacement, velocity, and acceleration of the ball on separate graphs for the first 5 seconds as a function of time for both part (a) and part (b). Make sure you mark vour axes d. Now write an expression for displacement, velocity, and acceleration of the ball for a CONSTANT acceleration of 10 cm/s and plot them on the same graphs. Make sure your TA can tell which is which What do you notice when you compare part (c) and part (d)?

Answer 1

a)

at t = 0, u = 10cm/s ; a = 10cm/s^2

s = ut+1/2 at^2 = 10x5 + 1/2 x 10 x5^2

= 50 + 125 = 175 cm

b)

same as above

c)

At t = 0, u = 10cm/s: a = 10cm/s^2 s = ut + 1/2 at^2 = 10x5 + 1/2 x 10 x5^2 = 50 + 125 = 175 cm same as above Starting from rest at position zero y = 1/2 at^2 v = at a = constant accelerating at 9.8 m/s^2 More generally y = y_0 + v_0 t + 1/2 at^2 v = v_0 + at Velocity is equal to the slope of the position curve. Acceleration is equal to the slope of the velocity curve. they will appear same

e)

they will appear same