StudentAidUnit 3: Dynamics — Numericals
9th Class Physics · Unit 3: Dynamics
3.1.A 10 kg block is placed on a smooth horizontal surface. A horizontal force of 5 N is applied to the block. Find: (a) the acceleration produced in the block. (b) the velocity of block after 5 seconds.
Given
Mass of block
m = 10 kg
Applied force
F = 5 N
Initial velocity
vi = 0 ms-1
Time
t = 5 s
Formula
Newton's second law
a = Fm
Working
(a) Acceleration
a = 510 = 0.5 ms-2
Formula
Kinematic equation for velocity
vf = vi + at
Working
(b) Final velocity
vf = 0 + (0.5)(5) = 2.5 ms-1
(a) 0.5 ms-2; (b) 2.5 ms-1
3.2.The mass of a person is 80 kg. What will be his weight on the Earth? What will be his weight on the Moon? The value of acceleration due to gravity of Moon is 1.6 m s².
Given
Mass of person
m = 80 kg
Gravitational acceleration on Earth
ge = 10 ms-2
Gravitational acceleration on Moon
gm = 1.6 ms-2
Formula
Weight formula
W = mg
Working
Weight on Earth
We = (80)(10) = 800 N
Weight on Moon
Wm = (80)(1.6) = 128 N
Weight on Earth = 800 N; Weight on Moon = 128 N
3.3.What force is required to increase the velocity of 800 kg car from 10 m s¹ to 30 m s¹ in 10 seconds?
Given
Mass of car
m = 800 kg
Initial velocity
vi = 10 ms-1
Final velocity
vf = 30 ms-1
Time
t = 10 s
Formula
Acceleration from kinematics
a = vf - vit
Working
a = 30 - 1010 = 2010 = 2 ms-2
Formula
Newton's second law
F = ma
Working
F = (800)(2) = 1600 N
1600 N
3.4.A 5g bullet is fired by a gun. The bullet moves with a velocity of 300 m s¹. If the mass of the gun is 10 kg, find the recoil speed of the gun.
Given
Mass of bullet
m = 5 g = 0.005 kg
Velocity of bullet
v = 300 ms-1
Mass of gun
M = 10 kg
Formula
Conservation of momentum (initially at rest)
m v + M V = 0
Rearrange for gun velocity
M V = -m v
V = -m vM
Working
V = -(0.005)(300)10 = -1.510 = -0.15 ms-1
-0.15 ms-1
3.5.An astronaut weighs 70 kg. He throws a wrench of mass 300 g at a speed of 3.5 m s¹. Determine: (a) the speed of astronaut as he recoils away from the wrench. (b) the distance covered by the astronaut in 30 minutes.
Given
Mass of astronaut
M = 70 kg
Mass of wrench
m = 300 g = 0.3 kg
Velocity of wrench
v = 3.5 ms-1
Time
t = 30 min = 1800 s
Formula
Conservation of momentum (recoil velocity)
V = -m vM
Working
(a) Astronaut recoil velocity
V = -(0.3)(3.5)70 = -1.0570 = -0.015 ms-1
Formula
Distance covered
S = |V| × t
Working
(b) Distance in 30 minutes
S = (0.015)(1800) = 27 m
(a) -0.015 ms-1; (b) 27 m
3.6.A 6.5 × 10³ kg bogie of a goods train is moving with a velocity of 0.8 m s¹. Another bogie of mass 9.2 × 10³ kg coming from behind with a velocity of 1.2 m s¹ collides with the first one and couples to it. Find the common velocity of the two bogies after they become coupled.
Given
Mass of bogie 1
m1 = 6.5 × 103 kg
Velocity of bogie 1
v1 = 0.8 ms-1
Mass of bogie 2
m2 = 9.2 × 103 kg
Velocity of bogie 2
v2 = 1.2 ms-1
Formula
Conservation of momentum (perfectly inelastic collision)
m1 v1 + m2 v2 = (m1 + m2) vf
Rearrange for final velocity
vf = m1 v1 + m2 v2m1 + m2
Working
vf = (6.5 × 103)(0.8) + (9.2 × 103)(1.2)(6.5 × 103) + (9.2 × 103)
vf = 5200 + 1104015700 = 1624015700
Result
Common velocity
vf = 1.03 ms-1
3.7.A cyclist weighing 55 kg rides a bicycle of mass 5 kg. He starts from rest and applies a force of 90 N for 8 seconds. Then he continues at a constant speed for another 8 seconds. Calculate the total distance travelled by the cyclist.
Given
Mass of cyclist
m1 = 55 kg
Mass of bicycle
m2 = 5 kg
Total mass
m = m1 + m2 = 60 kg
Applied force
F = 90 N
Acceleration phase duration
t1 = 8 s
Constant velocity phase duration
t2 = 8 s
Initial velocity
vi = 0 ms-1
Formula
Acceleration during phase 1
a = Fm = 9060 = 1.5 ms-2
Velocity at end of phase 1
vf = vi + at1 = 0 + (1.5)(8) = 12 ms-1
Distance in phase 1 (acceleration)
S1 = vi t1 + 12a t12
Working
S1 = (0)(8) + 12(1.5)(8)2 = 0.75 × 64 = 48 m
Formula
Distance in phase 2 (constant velocity)
S2 = vf × t2
Working
S2 = (12)(8) = 96 m
Total distance
S = S1 + S2 = 48 + 96 = 144 m
144 m
3.8.A ball of mass 0.4 kg is dropped on the floor from a height of 1.8 m. The ball rebounds straight upward to a height of 0.8 m. What is the magnitude and direction of the impulse applied to the ball by the floor?
Given
Mass of ball
m = 0.4 kg
Height before impact
h1 = 1.8 m
Height after rebound
h2 = 0.8 m
Acceleration due to gravity
g = 10 ms-2
Formula
Velocity before impact (downward)
vf2 = vi2 + 2 g h1
Working
vf2 = 0 + 2(10)(1.8) = 36
vf = 6 ms-1 (downward)
Formula
Velocity after rebound (upward from kinematics)
vi2 = vf2 + 2(-g) h2
Working
0 = vi2 + 2(-10)(0.8)
vi2 = 16
vi = 4 ms-1 (upward)
Formula
Impulse (change in momentum)
J = m(vf - vi) = m(vafter - vbefore)
Working
J = 0.4(4 - (-6)) = 0.4(10) = 4 Ns
4 Ns
3.9.Two balls of masses 0.2 kg and 0.4 kg are moving towards each other with velocities 20 m s¹ and 5 m s¹ respectively. After collision, the velocity of 0.2 kg ball becomes 6 m s¹. What will be the velocity of 0.4 kg ball?
Given
Mass of ball 1
m1 = 0.2 kg
Mass of ball 2
m2 = 0.4 kg
Velocity of ball 1 before collision
v1 = 20 ms-1
Velocity of ball 2 before collision (toward first ball)
v2 = -5 ms-1
Velocity of ball 1 after collision
v1' = 6 ms-1
Formula
Conservation of momentum
m1 v1 + m2 v2 = m1 v1' + m2 v2'
Working
(0.2)(20) + (0.4)(-5) = (0.2)(6) + (0.4) v2'
4 - 2 = 1.2 + (0.4) v2'
2 = 1.2 + (0.4) v2'
0.8 = (0.4) v2'
Result
Velocity of ball 2 after collision
v2' = 2 ms-1