CLASS 11 · PHYSICS · CHAPTER 4
Simple worked solutions to all 23 end-of-chapter questions on Newton’s laws, forces, momentum, friction and circular motion.
23 worked answers · short steps first, explanations when you need them
Textbook pages 68–70
Question 1
In (a)–(d), velocity is constant or the object is at rest, so acceleration is zero and forces balance. In (e), no force acts.
The net force is 0 N in all five cases. A zero net force has no direction.
Question 2
After release, gravity is the only force. Weight=mg=0.05×10=0.5 N downward.
It is 0.5 N downward in all three stages. A 45° launch does not change the force; at the top only the vertical velocity is zero, not necessarily the whole velocity.
Question 3
Once released, the train no longer pushes the stone. In (a)–(c), gravity is the only force: mg=0.1×10=1 N downward.
(a) 1 N down. (b) 1 N down. (c) 1 N down. (d) The stone accelerates with the train: horizontal force=ma=0.1×1=0.1 N forwards; vertical forces balance.
Question 4
Tension is the real horizontal force pointing towards the centre. The required centripetal force is the net inward force; it is not an extra force.
The correct answer is T (option i). In this motion T=mv²/l.
Question 5
Acceleration opposite motion: a=−F/m=−50/20=−2.5 m/s². Use v=u+at with v=0.
0=15−2.5t, so t=6 s.
Question 6
a=(v−u)/t=(3.5−2.0)/25=0.06 m/s². Since it speeds up in the same direction, force points along its motion.
F=ma=3.0×0.06=0.18 N, along the direction of motion.
Question 7
Resultant force magnitude=√(8²+6²)=10 N. Its angle θ from the 8 N direction has tanθ=6/8.
a=F/m=10/5=2 m/s². θ=tan⁻¹(6/8)≈37° from the 8 N force, towards the 6 N force.
Question 8
Convert speed: 36 km/h=10 m/s. Acceleration=(0−10)/4=−2.5 m/s². Total moving mass=400+65=465 kg.
F=ma=465(−2.5)=−1162.5 N. Average retarding-force magnitude≈1.16×10³ N, opposite the motion.
Question 9
Upward thrust must overcome weight and still provide upward acceleration: T−mg=ma.
T=m(g+a)=20,000(10+5)=3.0×10⁵ N upward.
Question 10
During the force, a=F/m=−8/0.40=−20 m/s². Before t=0 the body moves at 10 m/s. At t=30 s: x=10(30)+½(−20)(30²)=−8700 m and v=10−20(30)=−590 m/s.
At −5 s: x=10(−5)=−50 m. At 25 s: x=10(25)−10(25²)=−6000 m. At 100 s, no force acts after 30 s: x=−8700+(−590)(70)=−50,000 m.
Question 11
At release, stone’s horizontal speed matches the truck: vx=2×10=20 m/s. During the next second, gravity gives downward velocity vy=10 m/s. After release, no horizontal force acts.
Velocity=(20 m/s forward, 10 m/s downward); speed=√(20²+10²)=22.4 m/s, 26.6° below horizontal. Acceleration=10 m/s² downward.
Question 12
At an extreme point the bob is momentarily at rest. At the mean position its velocity is horizontal and tangent to the swing. After the cut only gravity acts.
(a) It falls vertically from rest. (b) It follows a projectile path: initially horizontal motion combined with downward acceleration, so the path is parabolic.
Question 13
The scale reads the normal force N. Take upward positive: N−mg=ma. Here mg=70×10=700 N.
(a) a=0: N=700 N. (b) a=−5: N=700−350=350 N. (c) a=+5: N=700+350=1050 N. (d) free fall a=−g: N=0.
Question 14
Each section is straight, so velocity is constant within that section and acceleration (hence force) is zero. The middle-section velocity is (3−0)/4=0.75 m/s; the velocity jumps from 0 to 0.75 at t=0, then back to 0 at t=4.
Force is 0 N in all three open time intervals. Impulse=mΔv: at t=0, J=4(0.75−0)=+3 N·s; at t=4, J=4(0−0.75)=−3 N·s.
Question 15
Both blocks accelerate together: a=F/(10+20)=600/30=20 m/s². Consider the block not directly pulled to find tension.
Question 16
The 12 kg mass goes down and the 8 kg mass goes up. For the pair, net driving force=(12−8)g and total mass=20 kg.
a=(12−8)10/(12+8)=2 m/s². For the 8 kg mass: T−80=8(2), so T=96 N.
Question 17
Initial total momentum is zero. If external forces are negligible during the short disintegration, momentum is conserved.
p₁+p₂=0, so p₂=−p₁. Their momenta have equal magnitudes and opposite directions; therefore the products move in opposite directions.
Question 18
For either ball, velocity reverses. Choose its initial direction positive: u=+6 m/s, v=−6 m/s. Change in momentum is m(v−u).
J=0.05(−6−6)=−0.60 N·s for that ball. Each impulse has magnitude 0.60 N·s; their directions are opposite.
Question 19
Before firing, total momentum is zero. After firing, gun and shell momenta must still add to zero: Mvgun+mshell vshell=0.
Gun recoil speed=(0.020×80)/100=0.016 m/s, opposite the shell.
Question 20
Speed=54×5/18=15 m/s. The initial and final momentum vectors have equal size mu=2.25 kg·m/s and are separated by 45°. The change vector is the chord between them: |Δp|=2mu sin(45°/2).
In the textbook’s deflection geometry, the component perpendicular to the deflection bisector reverses. Impulse=2mu cos22.5°=2(0.15)(15)cos22.5°≈4.16 N·s, along the change in momentum.
Question 21
Frequency=40/60=2/3 s⁻¹, so v=2πrf=2π(1.5)(2/3)=2π≈6.28 m/s. Tension supplies centripetal force T=mv²/r.
T=0.25(6.28²)/1.5≈6.58 N. At the limit, vmax=√(Tmax r/m)=√(200×1.5/0.25)=34.6 m/s.
Question 22
At the instant the string breaks, the inward tension disappears. The stone continues in the direction of its instantaneous velocity.
It flies off along the tangent to the circle at the break point (option b).
Question 23
(a) A horse moves forward by pushing the ground backward; the ground’s friction pushes it forward. In empty space there is no ground for this push. (b) The body tends to keep its motion when the bus stops. (c) Pulling upward at an angle reduces the normal force and friction; pushing downward increases them. (d) Increasing stopping time reduces force for the same momentum change.
(a) Without an external force from the ground, horse and cart cannot accelerate together. (b) Inertia carries passengers forward. (c) The reduced friction makes pulling easier. (d) A longer stopping time means a smaller force on the hands.