Q1. Explain how kinetic energy depends on mass and speed. Use examples to show the effect of doubling mass and doubling speed.
Answer:
Kinetic energy (KE) is the energy due to motion.
Its formula is E_k = 1/2 mv^2.
If you double the mass, KE also doubles (because KE ∝ m).
If you double the speed, KE becomes four times (because KE ∝ v^2).
Example: m = 2 kg, v = 3 m/s → KE = 1/2 × 2 × 9 = 9 J.
Double mass to 4 kg → KE = 18 J (doubles). Double speed to 6 m/s → KE = 1/2 × 2 × 36 = 36 J (four times).
Q2. In the ball-and-sand activity, explain why the dent is deeper when the ball is dropped from a greater height.
Answer:
A higher drop means larger potential energy (mgh) at the start.
As the ball falls, PE changes into kinetic energy.
So, greater height gives greater speed just before hitting the sand.
Larger speed means larger kinetic energy at impact.
The ball does more work on the sand, so the dent is deeper.
Thus, more height → more KE → more work done on the sand.
Q3. Derive the relation W = 1/2 m (v^2 − u^2) using motion equations and explain its meaning.
Answer:
Work done, W = F × s.
From Newton’s second law, F = m a.
So, W = m a s.
Use the kinematic relation: v^2 − u^2 = 2 a s.
Then, a s = (v^2 − u^2)/2.
Therefore, W = m × (v^2 − u^2)/2 = 1/2 m (v^2 − u^2), which equals the change in kinetic energy.
Q4. A trolley of 2 kg moves at 3 m/s and hits a block. If we make the trolley 4 kg but keep speed 3 m/s, how does the work done on the block change? Explain.
Answer:
Initial KE with 2 kg: KE1 = 1/2 × 2 × 3^2 = 9 J.
New KE with 4 kg: KE2 = 1/2 × 4 × 3^2 = 18 J.
The work done on the block equals the trolley’s KE (if it stops).
With double mass at the same speed, KE becomes double.
So the work on the block doubles, causing more displacement.
This shows KE is directly proportional to mass at fixed speed.
Q5. Why does a fast-moving bullet penetrate a target better than a slow object? Use kinetic energy ideas.
Answer:
Bullet has a very high speed.
KE depends on the square of speed (v^2).
Even with small mass, large speed gives huge KE.
On impact, this KE is used to do work inside the target.
More work means more penetration and deeper entry.
A slow object has much less KE, so it does less work and penetrates less.
High Complexity (Analysis & Scenario-Based)
Q6. Two identical balls are dropped into wet sand: one from 50 cm and one from 150 cm. Predict and explain the difference in the dents using energy ideas.
Answer:
At 50 cm: PE = m g h1. At 150 cm: PE = m g h2, where h2 = 3 × h1.
On falling, PE → kinetic energy at impact.
The ball from 150 cm has three times the PE, so three times the KE at impact.
It can do more work on the sand, so the dent is deeper.
Depth shows the energy transferred to the sand.
So, greater height → greater KE → greater dent.
Q7. A car increases speed from 30 km/h to 60 km/h. Explain how the work needed and the stopping distance change, if braking force remains the same.
Answer:
Convert speeds to m/s: 30 km/h ≈ 8.33 m/s, 60 km/h ≈ 16.67 m/s.
KE at 30 km/h: 1/2 m (8.33)^2. KE at 60 km/h: 1/2 m (16.67)^2.
KE becomes about four times when speed doubles.
Brakes do work: W = F × s to remove this KE.
If braking force is the same, distance s must be four times to remove four times the KE.
Hence, higher speed means much more work and longer stopping distance.
Q8. A heavy truck and a small car have the same kinetic energy. Compare their speeds and explain who is moving faster and why.
Answer:
KE formula: E_k = 1/2 m v^2.
Rearranged: v = sqrt(2 E_k / m).
For the same KE, speed is inversely related to sqrt(mass).
The truck has larger mass, so its speed is smaller.
The car has smaller mass, so its speed is larger.
Thus, same KE does not mean same speed; lighter objects move faster for the same KE.
Q9. In a trolley-block experiment, you increase the trolley’s speed by 50% while keeping its mass the same. Predict how the block’s displacement changes and explain using energy and work.
Answer:
Let initial speed be v; new speed is 1.5 v.
Initial KE: 1/2 m v^2. New KE: 1/2 m (1.5 v)^2 = 2.25 × (1/2 m v^2).
So KE becomes 2.25 times.
The work on the block equals the trolley’s KE (if it stops after impact).
With similar contact and friction, displacement is roughly proportional to work done.
Hence, the block’s displacement should be about 2.25 times larger.
Q10. Design a simple plan to verify W = ΔK in the lab using a trolley, weights, and a measured track. Explain the steps and reasoning.
Answer:
Pull a trolley with a small constant force using weights over a pulley.
Measure displacement s along a smooth track for a fixed pull.
Record initial speed u (start from rest for simplicity: u = 0).
Calculate work: W = F × s, where F equals the hanging weight (minus small friction).
Measure final speed v with a stopwatch and markers, or a photogate.
Compare W with ΔK = 1/2 m (v^2 − u^2). They should be approximately equal, confirming the work-energy theorem.
