# Some Physics questions you might find tricky

Let us start off with this question: Do you think that Physics is easy? You might want to reconsider this fallacy again by looking at the questions below:

**Question 1:**

A motorist traveling at 10m/s, was able to bring his car to rest in a distance of 10m. If he had been traveling at 30m/s, in what distance could he bring his cart to rest using the same breaking force?

**Explanation:**

You might think that the answer is 30m. This is wrong.

To comprehend this question, you need to sketch two speed-time graphs of the two cases, one for the first scenario and another for the next.

For the first scenario, in order to find the time taken to bring the car to rest in a distance of 10m, we let the time taken be *t* seconds. So, 1/2 x 10 x *t* = 10 → *t* = 2.

Then, we have to find the braking force needed.Letting the force be *F*, *F* = (0-10)/2 → *F = *-5N.

Subsequently, to find the time taken to bring the car to rest, which was traveling at 30m/s, we will form this equation: -5N = (0-30)/*t → t = *6.

Finally, the distance traveled by the car will be: D = 1/2 x 30 x 6 = **90m**. There you go.

**Question 2:**

In an experiment where the Brownian motion of smoke particles in air is studied, it is observed that the heavier particles settle at the bottom of the glass cell relatively quickly while the much smaller particles remain suspended in air for long periods of time. This is because______________.

**Explanation**:

A lot of people think that the gravitational force acting on the small particles is negligible. This is wrong. The reason should be that the random collisions with the air molecules keep the small particles suspended.

**Question 3**:

Lemonade at 25°C is cooled to 10°C by adding 100g of ice at 0°C. Given that the specific heat capacity of water and lemonade is 4.2 kJ kg^{-1} °C^{-1} and the specific latent heat of fusion of ice is 336 kJ kg^{-1 }, what is the mass of lemonade that was cooled?

**Explanation:**

We have to find the energy emanated from the ice. Letting the energy be *Q*, then *Q* = m x l_{f }= 0.1kg x 336 kJ kg^{-1} = 33.6 kJ.

Now, using the same amount of energy *Q*, *Q* = *mcΔθ → *33.6 kJ = m x 4.2 kJ kg^{-1} °C^{-1} x 15°C → m ≈ 533.0 g.

**Question 4:**

A ball is released from a height *h*, above a table. Given that air resistance is negligible and 50% of its kinetic energy is converted to other forms of energy at each bounce, what will be the height reached after the second bounce?

**Explanation:**

Sketching a diagram might help you. When the ball bounces the first time, 50% of its kinetic energy is converted to other forms of energy, denoting that the ball will only be able to reach a height of *h/*2* or *0.50*h* (half of the initial height). Now, for the second bounce, 50% of its kinetic energy is again converted to other forms of energy. This means that the ball can now only reach to a height of *h/*4 or **0.25 h** (half of 0.50h).

**Question 5:**

During a single lightning flash, 40 C of charge travels through the air in 20 ms. Given that the potential difference between the clouds and the earth is 2.0 x 10^{7} V, how much energy is dissipated during the lightning flash?

**Explanation:**

Use the formula *V = W/Q*, where V = 2.0 x 10^{7} V and Q = 40 C. Substitute these values into the formula and we will get W = 80 x 10^{7} J. If you want to leave the answer in standard form, it will be 8.0 x 10^{8} J. Note that the time = 20 ms is there just to confuse you. This is a ploy often used by teachers.

**Question 6:**

Estimate the minimum time it will take a 750W microwave oven to thaw 0.25kg of frozen soup. The soup is initially at -18°C and is to be just turned into liquid at 0°C. The soup can be assumed to be made entirely of water. Take the specific heat capacity of ice as 2100 J kg^{-1} °C^{-1} and the specific latent heat of fusion of water as 334 0000 J kg^{-1} .

**Explanation:**

For this question, we must use the formula **Q = mcΔθ + ml _{f}** (it is a common, tactless mistake of students to omit ml

_{f, }the latent heat of fusion of water. Take note in order for a solid to change its state into water, some energy is needed to overcome the forces of attraction between molecules.)

Pt = mcΔθ + ml_{f }(Q = Pt)

Substitute the appropriate values in.

750 t = 0.25 (2100 x 18 + 3340000)

**t = 1130s**

**Question 7:**

Explain, in terms of the behaviour of its molecules, why the pressure of the trapped air increases as its volume decreases.

**Explanation:**

You might think this is a simple question, and you will get full marks for this. Believe it or not, some students tend to overlook some key points in their answers, resulting in a loss of marks.

My suggested answer is this: When the volume of gas is decreased, the number of molecules per unit area is increased. So for every second, there is an increase in the number of collisions with each unit area of the container walls. The force exerting on the walls is increased, so the pressure increases.

**Question 8:**

A bullet of mass 0.020kg traveling horizontally at 50m/s is stopped after moving through a distance of 0.10m in a concrete block. What is the average retarding force applied to the bullet by the concrete?

**Explanation: **

v^{2} = u^{2} + 2as

0^{2} = 50^{2} + 2(a)(0.1)

a = -12500m/s^{2}

F = ma

= 0.020 x (-12500)

= -250N

The answer is 250N, not -250N. A retarding force is negative in connotation.

**Question 9: **

A bicycle travels up a hill at an average speed of 4m/s and down the same hill through the same distance at an average speed of 8m/s. Express the average speed for the entire journey in terms of *t* and *d*.

**Explanation: **

You have to speculate given this convoluted question. Let us assume that the time taken to travel up the hill is *t* seconds. Then, the time taken to travel down the hill is 1/2 of *t *seconds as the speed now is twice than that of the former. The total distance traversed, we assume, is 2d metres. Hence, the average speed is: **2d/(1/2 + 1)t. **There you go. You might have realized that this question isn't testing you on physics; rather, it is testing you on math.

**Question 10:**

During the flag raising assembly, the student pulls the rope with a force of 15N and the flag rises at a uniform speed of 0.5 m/min. Calculate the power supplied by the student.

**Explanation: **

There is nothing difficult in this question. It gives you a lucid description of the situation and the values are given conspicuously. All you need to do is to understand that

*F x s*= work done, that is, represented by the letter

*E*in the formula: P = E/t. The rest of the solution is given below:

P = E/t

= (F x s)/ t

= F x (s/t)

= F x speed

= 15N x 0.5 m/min

= 0.125W