Session 2: Kinematics Timed Practice Solutions

Timed Practice Solutions 1. Both balls will hit the ground with the same speed. Explanation: Descriptive Since air resistance is neglected, the ball experiences the same gravitational acceleration g = 9.81 ms-2 (acting downwards) on the way up and coming down. The first ball which was thrown upwards with the speed of u will decelerate at a rate of g until it reaches the maximum height where it stops instantaneously (v = 0 ms-1). Then it accelerates downwards with the same acceleration g. Hence, when it reaches the initial position from which it was thrown, it will have a speed of u but in the opposite direction (downwards). This was the velocity of the second ball when it was at the same position. It follows that when the ball hits the ground, it does so with the same velocity as the second ball. Or by Calculation (Note: the First ball

Second ball

Initial position

g

u u h

v ??

+ve

final displacement of both balls are the same. Final displacement = -h ) Taking the sign convention upwards as positive, Using

v2 = u2 + 2as ,

First ball:

v2 = u2 + 2 (-g)( -h)

v = - u 2  2gh Second ball:

v2 = (-u)2 + 2 (-g)( -h)

v ??

v = - u 2  2gh Hence both balls will hit the ground at the same speed v ms-1

2. u = 12 ms-1

g = 9.81 ms-2 80 m +ve

We are given a, u and s. We need to find t. (Note: initial speed u of the package is the speed of the balloon and not zero because the package is initially moving upwards with the balloon) So we use

1 2 at 2 1 80 = 12t +  9.81 t 2 2 9.81 2 t  12t  80  0 2 s = ut +

t = 5.44s or -3.00s (rejected)

3. v g

h +ve

θ

V

We resolve the final velocity V into the horizontal component Vx and vertical component Vy. Note that the angle θ is largest when Vx is minimum and Vy is maximum.

Vx V θ For Vx :

Vy

Assuming no air resistance, Vx = v (where v is the initial horizontal velocity shown in the diagram) Even if there is air resistance, Vx will also be smaller if the initial horizontal velocity v is smaller since there is a retardation in the horizontal motion. So we choose the smallest value of v given to ensure that Vx is also smallest. For Vy: (Refer to first diagram) For the vertical motion, since we are given a, u and s, and we need to find v, We will use v2 = u2 + 2as Vy2 = 02 + 2(-9.81)(-h) Vy2 = 2(9.81)h Hence, Vy is maximum if h is maximum. So we will choose the maximum value of h given if we want Vy to be maximum Ans. B (minimum v, maximum h given)