Our Dynamic Universe

Question 1

Answer 1

1.) C.

Since the velocity decreases uniformly, the acceleration must be constant. The only force acting on the ball (ignoring air resistance) is gravity, which provides a constant acceleration of -9.8 m/s^2 throughout the motion.

Question 2

Answer 2

2.) B
s = ut + 1/2 at^2
s = (12 x 6) + 1/2 x (-1.5) x (6^2)
s = 45m

3.) A
For an object of mass on an inclined plane at angle (theta),
The component of weight acting DOWN the slope is - mg sin (theta)
The component of weight acting NORMAL to the slope is mg cos (theta)

4.) D
The scales show a reading greater than the persons weight, which means the normal force is greater than the mg.
This occurs when the lift either accelerates upwards or decelerates while moving downwards.

Question 3

Answer 3

5.) C
F = ma
F = 400 x 2.0
F = 800 N

6.) B
Ep = mgh
Where:
m = 6.25 x 10^8 / 60 = 1.04x10^7 kg/s (10416666.67) kg/s
g = 9.8 m/s^2
h = 108m
Ep = 1.04x10^7 x 9.8 x 108
Ep = 1.10 x 10^10 W

7.) C
L’ = L √1- ( v/c) ) squared
190 = L √ 1- (0.60) squared
190 = L x 0.8
190 / 0.8 = L
L = 238m

Question 4

Answer 4

8.) E

fo = fs ( v / v +- vs )
fo = 750 ( 340 / 340 - 25 )
fo = 810 hz

Question 5

Answer 5

2.)
m1 u1 + m2 u2 = m1 v1 + m2 v2

(0.25 x 1.2) + (0.45 x -0.6) = (0.25 x -0.8) + ( 0.45 x v2)

= (0.3) + (-0.27) = (-0.2) + (0.45 v2)

0.03 = -0.2 + 0.45v2

0.03 + 0.2 = 0.45v2

0.23 = 0.45v2

v2 = 0.23/0.45

v2 = 0.51 m/s

Question 6

Answer 6

2bi.) impulse is calculated as area under the force time graph

Area = 1/2 x Base x Height
Area = 1/2 x 0.25 x 4
Impulse = 0.5 N s

ii.) 0.5 kg ms^-1

Question 7

Answer 7