Let Us Assess, Q1–Q9 – Solutions വിലയിരുത്താം
The first nine end-of-chapter questions: a body braking from 144 km/h, a hockey ball that bounces back, a slowing lorry, the odd one out, the coin on a card, beating a carpet, the horse and the cart, a friction force read from Graph 3.1, and a force that doubles a velocity. Try each one in your notebook first, then step through the solution.
- Momentum p = mv (unit kg m/s). Change in momentum Δp = mv − mu. Rate of change of momentum = m(v − u)/t (unit kg m/s² = N).
- Second law: F = m(v − u)/t = ma. Acceleration a = (v − u)/t. Convert km/h to m/s with × 5/18 and g to kg with ÷ 1000.
- Choose one direction as positive (usually the first direction of motion); anything opposite gets a minus sign.
- Inertia of rest / of motion. Third law: action and reaction are equal and opposite and act on different bodies.
Question 1 · Braking from 144 km/h
1. 144 km/h പ്രവേഗത്തിൽ സഞ്ചരിക്കുന്ന 5 kg മാസുള്ള ഒരു വസ്തു 4 s കൊണ്ട് നിശ്ചലാവസ്ഥയിലാകുന്നുവെങ്കിൽ താഴെക്കൊടുത്തവ കണക്കാക്കുക. വസ്തുവിന്റെ a) ആദ്യമൊമെന്റം b) അന്ത്യമൊമെന്റം c) മൊമെന്റവ്യത്യാസം d) മൊമെന്റവ്യത്യാസനിരക്ക് An object of mass 5 kg moving with a velocity of 144 km/h comes to rest in 4 s. Calculate the object's
- initial momentum ആദ്യമൊമെന്റം
- final momentum അന്ത്യമൊമെന്റം
- change in momentum മൊമെന്റവ്യത്യാസം
- rate of change of momentum മൊമെന്റവ്യത്യാസനിരക്ക്
First change 144 km/h into m/s (× 5/18). “Comes to rest” means v = 0. Then p = mv, Δp = mv − mu, rate = Δp ÷ t.
The speeds on the slider are multiples of 18 km/h, so they convert to whole numbers of m/s (18 km/h = 5 m/s).
(a) initial momentum mu = 5 × 40 = 200 kg m/s
(b) final momentum mv = 5 × 0 = 0
(c) change in momentum mv − mu = 0 − 200 = −200 kg m/s
(d) rate of change of momentum = −200 ÷ 4 = −50 kg m/s² (a force of 50 N acting opposite to the motion).
Using 144 directly: 5 × 144 = 720 is wrong, because the unit kg m/s needs the speed in m/s. And Δp is final − initial, so it comes out negative here – keep the minus sign and say what it means.
Question 2 · The hockey ball that bounces back
2. 200 g മാസുള്ള ഒരു ഹോക്കി ബോൾ 20 m/s വേഗത്തിൽ ഒരു ഹോക്കിസ്റ്റിക്കിൽ വന്നിടിച്ച് അതേ വേഗത്തിൽ അതേ പാതയിലൂടെ തിരികെ പോകുന്നു. ഈ ബോളിന്റെ മൊമെന്റവ്യത്യാസം എത്ര? A hockey ball of mass 200 g strikes a hockey stick at a speed of 20 m/s and goes back along the same path with the same speed. What is the change in momentum of the ball?
Same speed – but not the same velocity! Take the direction towards the stick as +. Then u = +20 m/s and v = −20 m/s. Don't forget 200 g = 0.2 kg.
Set the rebound speed to 0: the ball stops dead on the stick and Δp = −4 kg m/s. Bouncing back at the same speed doubles the change to −8 kg m/s – so the stick must push twice as hard.
Δp = mv − mu = 0.2 × (−20) − 0.2 × 20 = −4 − 4 = −8 kg m/s.
The size of the change is 8 kg m/s; the minus sign says it points away from the stick (opposite to the ball's first motion).
Answering “0, because the speed is the same”. Momentum is a vector: +4 kg m/s turned into −4 kg m/s, a change of −8 kg m/s. For any ball that bounces straight back at the same speed, Δp = −2mu.
A bounce needs twice the change in momentum of a dead stop. That is why a ball that bounces off a bat stings more than one that is caught and stopped, and why a cricket ball smashed back past the bowler needs such a big impulse from the bat.
Question 3 · A loaded lorry slows down
3. ഭാരം നിറച്ച 10,000 kg മാസുള്ള ഒരു ലോറിയുടെ പ്രവേഗം 4 s കൊണ്ട് 15 m/s ൽ നിന്നും 12 m/s ആയി മാറുന്നുവെങ്കിൽ ലോറിയുടെ മൊമെന്റവ്യത്യാസനിരക്കെത്ര? The velocity of a loaded lorry of mass 10,000 kg changes from 15 m/s to 12 m/s in 4 s. What is the rate of change of momentum of the lorry?
Rate of change of momentum = m(v − u)/t. Here v is smaller than u, so expect a negative answer.
Change in momentum = m(v − u) = 10000 × (12 − 15) = −30000 kg m/s.
Rate of change of momentum = −30000 ÷ 4 = −7500 kg m/s² (= −7500 N: a force of 7500 N acting against the motion).
The lorry lost only 3 m/s, yet the force is 7500 N – because the mass is huge. A heavily loaded lorry or KSRTC bus needs far more braking force (or far more time and distance) to slow down than a car at the same speed.
Question 4 · The odd one out
4. കൂട്ടത്തിൽപ്പെടാത്തതേത്? (ബലം, മൊമെന്റം, പ്രവേഗം, വേഗം) Which is the odd one out? (force, momentum, velocity, speed)
For each quantity ask: does it need a direction to be described completely?
Force, momentum and velocity are vector quantities (സദിശ അളവുകൾ) – each has a size and a direction (a push to the east is not the same as a push to the west; momentum points the same way as velocity). Speed has only a size: “40 km/h” is complete without a direction, so it is a scalar.
Question 5 · The coin on the card
5. ഒരു കപ്പിനു മുകളിൽ ഒരു കാർഡ്ബോർഡും അതിനു മുകളിൽ ഒരു നാണയവും വച്ചിരിക്കുന്നു. a) കാർഡ്ബോർഡിനെ പെട്ടെന്ന് തട്ടിത്തെറിപ്പിച്ചാൽ നാണയത്തിന് എന്തു സംഭവിക്കും? b) കാരണമെന്ത്? A piece of cardboard is placed on top of a cup and a coin is placed on the cardboard. (a) If the cardboard is flicked away suddenly, what will happen to the coin? (b) What is the reason?
Before the flick the coin is at rest. Is any horizontal force able to act on the coin for long enough to move it?
(b) Inertia of rest (നിശ്ചലജഡത്വം). The coin was at rest and tends to stay at rest. The flick acts on the card, not on the coin; the card slides away so quickly that the small friction between card and coin has almost no time to move the coin. With its support gone, the coin falls into the cup under gravity.
Pull slowly and friction has plenty of time: the coin is dragged along with the card and lands outside the cup. The trick works only because the force on the coin acts for a very short time.
Question 6 · Beating a carpet
6. ഒരു കാർപ്പറ്റ് വൃത്തിയാക്കാൻ വേണ്ടി നാം അതിനെ തൂക്കിപ്പിടിച്ച് അതിൽ കമ്പ് കൊണ്ട് അടിക്കാറുണ്ട്. ഇപ്രകാരം ചെയ്യുമ്പോൾ പൊടി വേർപെടുന്നു. ഇതിന്റെ കാരണം വ്യക്തമാക്കുക. To clean a carpet we hang it up and beat it with a stick. When we do this, the dust separates out. Explain the reason.
The stick hits the carpet, not the dust. Which one starts moving suddenly, and which one “wants” to stay where it was?
Shaking a mango-tree branch so that the mangoes drop, snapping a wet towel, or tapping a tiffin box upside down to loosen rice – each time the support moves suddenly while the object tends to stay at rest.
Question 7 · The horse and the cart
7. ഒരു കുതിരവണ്ടിയെ കുതിര വലിക്കുമ്പോൾ വണ്ടി മുന്നോട്ടു പോകുന്നു. അതേ സമയം കുതിരവണ്ടി കുതിരയെ എതിർദിശയിൽ തുല്യബലത്തോടെ വലിക്കും. പക്ഷേ കുതിരയും വണ്ടിയും മുന്നോട്ട് പോകും. ഇതെങ്ങനെ സാധ്യമാകുന്നു എന്ന് വിശദീകരിക്കുക. When a horse pulls a horse-cart, the cart moves forward. At the same time the cart pulls the horse in the opposite direction with an equal force. But the horse and the cart both move forward. Explain how this is possible.
Draw the forces on the cart alone, then on the horse alone. The two equal pulls are on different bodies. What pushes the horse forward? (Think of what its hooves do to the ground.)
Forward (to the right) is positive. The numbers are an example (horse 400 kg, cart 300 kg) chosen to show the idea; the textbook gives none.
• On the cart: the horse's forward pull is greater than the friction from the road on the wheels, so the cart has an unbalanced forward force and moves forward.
• On the horse: the horse pushes the ground backwards with its hooves; by the third law the ground pushes the horse forwards. This forward push is greater than the backward pull of the cart, so the horse also has an unbalanced forward force and moves forward.
So both move forward together.
“Equal and opposite forces cancel, so nothing should move.” Forces cancel only when they act on the same body. Never add an action to its reaction.
Question 8 · Friction from a velocity–time graph
8. 250 g മാസുള്ള വസ്തു ഒരു പ്രതലത്തിലൂടെ ചലിക്കുന്നതിന്റെ പ്രവേഗ - സമയ ഗ്രാഫ് നൽകിയിരിക്കുന്നു. വസ്തുവിൽ പ്രതലം പ്രയോഗിച്ച ഘർഷണബലം കണക്കാക്കുക. The velocity–time graph (Graph 3.1) of an object of mass 250 g moving on a surface is given. Calculate the frictional force (ഘർഷണബലം) exerted by the surface on the object.
Read u and v and the time from the graph. The slope of a v–t graph is the acceleration. Then F = ma with m in kg. Friction is the only horizontal force, so F is the friction.
A steeper line means a bigger retardation and so a bigger friction force. The red arrow (friction) stays the same size all the way, because the line is straight.
a = (v − u)/t = (0 − 6)/12 = −0.5 m/s²
F = ma = 0.25 × (−0.5) = −0.125 N.
So the surface exerts a frictional force of 0.125 N opposite to the direction of motion.
Using m = 250 (forgetting to change grams into kilograms) gives 125 N – a thousand times too big. And a downward-sloping v–t line means negative acceleration: report the friction as 0.125 N against the motion.
Question 9 · Doubling the velocity
9. 500 g മാസുള്ള ഒരു വസ്തു 40 m/s പ്രവേഗത്തിൽ ചലിക്കുന്നു. 4 s സമയത്തേക്ക് ഒരു ബലം പ്രയോഗിച്ചപ്പോൾ പ്രവേഗം 80 m/s ആയി മാറിയെങ്കിൽ പ്രയോഗിച്ച ബലം കണക്കാക്കുക. An object of mass 500 g is moving with a velocity of 40 m/s. When a force was applied for 4 s, its velocity became 80 m/s. Calculate the force applied.
F = m(v − u)/t. Change 500 g to kg first.
F = m(v − u)/t = 0.5 × (80 − 40)/4 = 0.5 × 40/4 = 20/4 = 5 N, in the direction of motion.
(Check: a = (80 − 40)/4 = 10 m/s², and F = ma = 0.5 × 10 = 5 N ✓.)
Positive answer, positive meaning: the force acts along the motion and speeds the object up. Compare Q1, Q3 and Q8, where the answers were negative because the force acted against the motion.