Kinetic Energy — Long Answer Questions (Class 9 Science, Physics)
Medium Level (Application & Explanation)
Q1. Explain, with
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to the sand-drop activity, how the height from which a ball is dropped affects the dent made in the sand. Relate this to the formula of kinetic energy.
Answer:
When a ball is dropped from a height it gains speed due to gravity; a greater height gives a higher final velocity when it reaches the sand.
According to the kinetic energy formula, E_k = 1/2 mv^2, kinetic energy depends on the square of velocity. So when velocity increases, energy increases rapidly.
The extra kinetic energy is transferred to the sand upon impact and used to displace sand grains, producing a deeper dent.
In the activity, drops from 25 cm, 50 cm, 1 m and 1.5 m showed increasing dent depth because each larger height gave the ball more energy on impact.
This experiment demonstrates energy conversion: gravitational potential energy → kinetic energy → work done on sand. The deeper dent is direct evidence of greater energy transfer.
Q2. A 15 kg object moves at 4 m/s. Describe step-by-step how to calculate its kinetic energy and explain what this value means in practical terms.
Answer:
Step 1: Identify mass m = 15 kg and velocity v = 4 m/s.
Step 2: Use the formula E_k = 1/2 mv^2. Plug values: E_k = 0.5 × 15 × 4^2.
Step 3: Compute 4^2 = 16, then 0.5 × 15 = 7.5, so E_k = 7.5 × 16 = 120 J.
Practical meaning: The object has 120 joules of energy due to motion. If it collides with another body, this energy can do work — for example, deforming the object, moving another object, or producing heat and sound.
Knowing kinetic energy helps predict damage potential, stopping forces required, and energy transformations during collisions or braking.
Q3. Using the trolley-and-block activity, explain why increasing the trolley’s mass or speed changes how far the block moves. Link your answer to work done and energy transfer.
Answer:
The trolley carries kinetic energy equal to 1/2 mv^2. When it hits the block, part of this energy is transferred to the block as work, causing the block to move.
Increasing the trolley’s mass (m) or its speed (v) increases the trolley’s kinetic energy. Because velocity is squared, increasing speed has a stronger effect than the same proportional increase in mass.
The work done on the block equals the energy transferred: W = F·s, where s is block displacement. If more energy is available, the block travels farther against friction and resistance.
Thus the activity shows how energy amount controls resulting displacement, illustrating the connection between kinetic energy and mechanical work.
Q4. Describe how the work-energy relation W = 1/2 m(v^2 − u^2) is used to find work done when a car accelerates. Give a clear example using simple numbers.
Answer:
The relation W = 1/2 m (v^2 − u^2) gives the work done to change an object’s speed from u (initial) to v (final). It comes from combining work with the change in kinetic energy.
For example, a 1000 kg car accelerates from u = 10 m/s to v = 20 m/s. Compute squared speeds: 10^2 = 100, 20^2 = 400. Difference = 300.
This means the engine (or net force) supplied 150 kJ of work to increase the car’s kinetic energy. In practice some energy also goes to friction and heat, so fuel consumption will be higher than this ideal value.
Q5. Why does doubling the velocity of an object quadruple its kinetic energy? Use algebraic reasoning and a simple explanation suitable for Class 9 students.
Answer:
Start from E_k = 1/2 mv^2. If velocity v is doubled to 2v, substitute: E'_k = 1/2 m (2v)^2.
Calculate square: (2v)^2 = 4v^2, so E'_k = 1/2 m × 4v^2 = 4 × (1/2 m v^2) = 4 E_k.
So doubling speed multiplies kinetic energy by four.
Simple explanation: speed affects energy much more strongly than mass because velocity is squared. Therefore a small increase in speed produces a large increase in energy, which is why high speeds are dangerous — stopping requires removing a lot more energy.
High Complexity (Analytical & Scenario-Based)
Q6. A car of mass 1500 kg increases speed from 30 km/h to 60 km/h. Calculate the work needed for this change. Show conversions and explain why the work is not simply doubled even though speed is doubled.
Answer:
Convert speeds to m/s: 30 km/h = 30×1000/3600 = 8.33 m/s (≈ 25/3 m/s) and 60 km/h = 16.67 m/s (≈ 50/3 m/s).
Use W = 1/2 m (v^2 − u^2). Compute squares: u^2 ≈ 8.33^2 = 69.44, v^2 ≈ 16.67^2 = 277.78. Difference ≈ 208.34.
Explanation: Although speed doubled, kinetic energy increased by a factor of four (since E_k ∝ v^2), so the work required is not just double. Increasing speed from 30 to 60 km/h needs three times more energy than increasing from 0 to 30, showing how energy rises fast with speed and why accelerating to high speeds consumes much more fuel/energy.
Q7. Design a simple experiment (using the sand-drop idea) to estimate the relative kinetic energies of two balls of different masses dropped from the same height. Describe procedure, observations, and how to interpret results.
Answer:
Procedure: Use two balls with different masses but similar shape and size (to reduce air resistance). Drop each from the same fixed height onto a thick, uniform bed of wet sand. Repeat several times for each ball and measure the dent depth for each drop. Keep dropping location and sand condition identical.
Observations: Heavier ball should make a deeper dent than lighter ball if both hit at same speed, indicating more energy transfer. Record average depth for each ball.
Interpretation: Since both balls have same velocity at impact (dropped from same height), kinetic energy difference comes from mass (E_k ∝ m). Deeper dent means greater work done on sand, so the dent depth is proportional to kinetic energy. Use depth ratios to compare relative energies. Note possible errors: different shapes, sand packing, and energy lost to sound/heat.
Q8. Consider a bullet and a slowly moving hammer hitting a wooden target. Explain, using kinetic energy and time of impact concepts, why the bullet penetrates more despite having possibly similar mass.
Answer:
Penetration depends on energy delivered and the rate of delivery (power/force over short time). A bullet has very high velocity, so its kinetic energy 1/2 mv^2 is large even for small mass.
On impact the bullet's energy is released over a very short time and small area, producing a very large force (force ≈ change in momentum / time). Shorter impact time means higher peak force, causing penetration.
A slowly moving hammer may have similar or even greater mass but much smaller velocity, so its kinetic energy and impact force are lower; the energy is distributed over a larger area and longer time, causing crushing or denting rather than penetration.
Thus both magnitude of kinetic energy and impact duration/area determine penetration ability.
Q9. A student argues that braking distance of a vehicle depends more on mass than on speed. Evaluate this statement using kinetic energy concepts and state what actually affects braking distance.
Answer:
Braking distance is determined by converting the vehicle’s kinetic energy into work done against braking force and friction. Since E_k = 1/2 mv^2, braking distance depends on both mass (m) and square of speed (v^2).
If braking force (friction) is roughly constant, distance ∝ E_k / force ∝ (1/2 mv^2)/F. So for fixed braking force, doubling mass doubles required work and roughly doubles distance, but doubling speed quadruples kinetic energy and so roughly quadruples braking distance.
Therefore speed has a stronger effect (because of the square). Other factors affecting braking distance include road conditions, brake efficiency, tire grip, and reaction time. The student’s claim is incorrect: speed affects braking distance more dramatically than mass.
Q10. Explain energy transformations and losses when a trolley collides with a wooden block and comes to rest. Where does the trolley’s initial kinetic energy go? Discuss in detail.
Answer:
Before collision, the trolley has kinetic energy (1/2 mv^2). During impact this energy is partly transferred to the block as kinetic energy, causing block movement and displacement (work done).
Other portions of energy convert into thermal energy due to friction between surfaces and internal deformation (heating of trolley, block, and contact surfaces). Some energy becomes sound energy (the collision noise). If deformation occurs, energy is used for plastic deformation (permanent shape changes).
If the trolley comes to rest, its initial kinetic energy has been fully converted into these forms: work on block, heat, sound, and internal energy. The exact distribution depends on elasticity of collision: an inelastic collision converts more into heat and deformation, while an elastic one returns more energy into motion (less converted to heat). This explains why collisions often produce permanent damage and heat.
