Exploration - Science solution
Class 9 - Chapter 6: How Forces Affect Motion
Exercise Solution
1. A table is moved with constant velocity
When a table moves with a constant velocity, its acceleration is zero.
According to Newton's First Law of Motion, the net force acting on the table must be zero.
Therefore,
Applied Force = Frictional Force
Answer: The frictional force exerted by the floor is equal in magnitude and opposite in direction to the applied force F.
2. Ball moving on a smooth frictionless surface
(i)
If no net force is applied on the ball, its velocity remains the same.
Answer: Same
(ii)
If a net force acts in the direction of motion, the ball accelerates and its speed increases.
Answer: Increase
(iii)
If a net force acts opposite to the direction of motion, the speed decreases.
Answer: Decrease
3. Blocks P and Q
For block P:
Net force = 5 N − 4 N = 1 N
Therefore, P experiences a net force.
For block Q:
Q moves with constant velocity.
Hence acceleration = 0.
Net force = 0.
Correct option: (i)
P experiences a net force and Q does not experience a net force.
4. Snake Boat Race
Total rowers = 100
Rowers rowing correctly = 95
Rowers rowing opposite = 5
Force by one rower = 200 N
Forward force = 95 × 200
= 19000 N
Backward force = 5 × 200
= 1000 N
Net force = 19000 − 1000
= 18000 N
Answer: Net force on the boat = 18000 N forward.
5. When a net force acts on an object
According to Newton's Second Law:
Acceleration is produced in the direction of the applied force and is directly proportional to the force.
Correct option: (iv)
6. Net force acts on which object?
Object A
The graph is a straight line with constant slope.
Velocity is constant.
Acceleration = 0.
Net force = 0.
Object B
Position remains constant.
Velocity = 0.
Acceleration = 0.
Net force = 0.
Object C
The graph is curved and slope increases continuously.
Velocity changes with time.
Acceleration is present.
Net force acts on this object.
Object D
Straight line with constant negative slope.
Constant velocity.
Acceleration = 0.
Net force = 0.
Answer: Net force acts on Object C.
7. Sailor jumping from a boat
Yes, the boat will move.
When the sailor jumps forward, he pushes the boat backward.
According to Newton's Third Law of Motion, every action has an equal and opposite reaction.
Therefore, the boat moves backward while the sailor moves forward.
Answer: The boat moves backward due to the reaction force exerted by the sailor.
8. Why is a landing mat or sand bed used in high jump?
When an athlete lands on a soft mat or sand bed, the stopping time increases.
From Newton's Second Law:
Force = Change in Momentum / Time
A larger stopping time results in a smaller impact force.
This reduces the chances of injury.
Answer: A landing mat or sand bed increases the time taken to stop, thereby reducing the impact force and preventing injuries.
9. Handcart Collision
According to Newton's Third Law of Motion, whenever two objects interact, they exert equal and opposite forces on each other.
Therefore, during the collision:
Force exerted by loaded cart on empty cart = Force exerted by empty cart on loaded cart
Correct Option: (iv)
The loaded cart and the empty cart both exert equal magnitude of force on each other.
10. Force-Mass Graph
From Fig. 6.40:
Mass (m) and acceleration (a) are inversely proportional.
Sample values from the graph:
- m = 1 kg, a = 10 m s-2
- m = 2 kg, a = 5 m s-2
- m = 5 kg, a = 2 m s-2
Using Newton's Second Law:
F = ma
F = 1 × 10 = 10 N
F = 2 × 5 = 10 N
F = 5 × 2 = 10 N
Thus the force remains constant at 10 N.
Answer: The Force–Mass graph will be a horizontal straight line parallel to the mass axis at F = 10 N.
11. Acceleration from Velocity-Time Graph
From Fig. 6.41:
Initial velocity, u = 10 m s-1
Final velocity, v = 30 m s-1
Time, t = 8 s
Acceleration,
a = (v − u)/t
a = (30 − 10)/8
a = 20/8
a = 2.5 m s-2
Answer: Acceleration = 2.5 m s-2
12. Bullet Entering Wooden Block
Mass of bullet,
m = 50 g = 0.05 kg
Initial velocity,
u = 100 m s-1
Final velocity,
v = 0
Distance travelled,
s = 50 cm = 0.5 m
Using:
v2 = u2 + 2as
0 = 1002 + 2 × a × 0.5
0 = 10000 + a
a = −10000 m s-2
Using:
F = ma
F = 0.05 × (−10000)
F = −500 N
Magnitude of force = 500 N
Answer: Stopping force on the bullet = 500 N
13. Football Kicked by a Footballer
Speed,
v = 108 km h-1
= 108 × 5/18
= 30 m s-1
Mass,
m = 0.4 kg
Force,
F = 800 N
Initially the ball is at rest.
u = 0
Using:
F = ma
800 = 0.4a
a = 2000 m s-2
Using:
v = u + at
30 = 0 + 2000t
t = 30/2000
t = 0.015 s
Answer: Time of contact = 0.015 s
14. Motion on a Rough Patch
Mass,
m = 2 kg
Initial velocity,
u = 10 m s-1
Friction force = 7 N
Applied force = 3 N
Net retarding force:
F = 7 − 3 = 4 N
Acceleration:
a = F/m
a = −4/2
a = −2 m s-2
At stopping:
v = 0
Using:
v2 = u2 + 2as
0 = 102 + 2(−2)s
0 = 100 − 4s
s = 25 m
Answer: The object travels 25 m before coming to rest.
15. Tractor Pulling Harrow and Trolley
For the harrow:
F = m1a1
m1 = F/a1
For the trolley:
F = m2a2
m2 = F/a2
Combined mass:
M = m1 + m2
M = F/a1 + F/a2
M = F[(a1 + a2)/(a1a2)]
Resulting acceleration:
a = F/M
a = F ÷ {F[(a1 + a2)/(a1a2)]}
a = (a1a2)/(a1 + a2)
Answer:
a = (a1a2)/(a1 + a2)
16. Bar Magnet and Magnetic Compass
The magnetic force on the compass needle and the bar magnet are equal and opposite, according to Newton's Third Law.
However, the compass needle has very small mass and is free to rotate.
Therefore, even a small magnetic force can produce noticeable acceleration and rotation.
The bar magnet has much larger mass and usually remains fixed in position.
Its acceleration is extremely small and cannot be noticed.
Answer: The compass needle moves because it has very small mass and can rotate freely, whereas the bar magnet has much larger mass and therefore shows negligible motion despite experiencing an equal magnetic force.