Set 4 k

Q1 [3 Marks]: State the condition under which a falling body experiences 'free fall'. Does the acceleration due to gravity depend on the mass of the falling body?

Part A: Condition for Free Fall
A body is said to be in 'free fall' when the only force acting upon it is the gravitational pull of the Earth. This means that all other forces, such as air resistance (drag) or buoyant forces, must be completely negligible or zero.

Part B: Dependence on Mass
No, the acceleration due to gravity ($g$) does not depend on the mass of the falling body.

Justification: According to Newton's Law of Universal Gravitation, the force on a body of mass $m$ is $F = \frac{GmM}{R^2}$. By Newton's Second Law, $F = mg$. Equating them gives $mg = \frac{GmM}{R^2}$, which simplifies to $g = \frac{GM}{R^2}$. The mass $m$ of the falling object cancels out completely.

Q2 [3 Marks]: A ball is thrown vertically upwards with a velocity of $20\text{ m/s}$. Taking $g = 10\text{ m/s}^2$, calculate the maximum height reached and the total time of flight.

Part A: Maximum Height
Step 1: Identify the given values. Initial velocity $u = +20\text{ m/s}$, acceleration $a = -g = -10\text{ m/s}^2$. At the maximum height, final velocity $v = 0$.

Step 2: Use the third equation of motion: $v^2 = u^2 + 2as$.

$$0^2 = (20)^2 + 2(-10)H_{max}$$

$$0 = 400 - 20H_{max}$$

$$20H_{max} = 400 \Rightarrow H_{max} = \frac{400}{20}$$

$$\text{Maximum Height } (H_{max}) = 20\text{ m}$$

Part B: Total Time of Flight
Step 3: Time to reach maximum height ($t_{up}$) can be found using $v = u + at$.

$$0 = 20 - 10t_{up} \Rightarrow 10t_{up} = 20 \Rightarrow t_{up} = 2\text{ s}$$

Step 4: The total time of flight ($T$) is twice the time of ascent (since air resistance is ignored).

$$T = 2 \times t_{up} = 2 \times 2$$

$$\text{Total Time of Flight } (T) = 4\text{ s}$$

Q3 [4 Marks]: A stone is dropped from a tower $100\text{ m}$ high. Simultaneously, another stone is projected vertically upwards from the ground with a velocity of $25\text{ m/s}$. When and where will they meet?

Step 1: Set up the equations for both stones.
Let them meet after time $t$ at a distance $y$ from the top of the tower.
For simplicity, let's take $g = 10\text{ m/s}^2$.

Step 2: Equation for Stone 1 (Dropped from top)
Initial velocity $u = 0$, distance fallen $= y$.

$$y = \frac{1}{2}gt^2 = \frac{1}{2}(10)t^2 = 5t^2 \quad \text{--- (Equation 1)}$$

Step 3: Equation for Stone 2 (Thrown from ground)
Initial velocity $u = 25\text{ m/s}$. The distance it travels upwards is $(100 - y)$.

$$100 - y = ut - \frac{1}{2}gt^2$$

$$100 - y = 25t - 5t^2 \quad \text{--- (Equation 2)}$$

Step 4: Solve for time ($t$).
Substitute $y = 5t^2$ from Equation 1 into Equation 2.

$$100 - 5t^2 = 25t - 5t^2$$

The $-5t^2$ terms cancel out on both sides.

$$100 = 25t \Rightarrow t = \frac{100}{25}$$

$$\text{They will meet at } t = 4\text{ s}$$

Step 5: Solve for position.
Substitute $t = 4\text{ s}$ back into Equation 1.

$$y = 5(4)^2 = 5(16) = 80\text{ m}$$

$$\text{They meet } 80\text{ m below the top (or } 20\text{ m above the ground)}$$

Q4 [5 Marks]: A balloon is ascending at a constant rate of $10\text{ m/s}$. At a height of $40\text{ m}$ above the ground, a food packet is dropped from it. Find the time taken by the packet to reach the ground and its velocity upon impact. ($g = 9.8\text{ m/s}^2$).

Crucial Concept: When an object is dropped from a moving body, it inherits the velocity of that body. The packet does NOT start with $u=0$. It starts with an initial upward velocity of $10\text{ m/s}$.

Step 1: Assign proper sign conventions.
Let upward be positive (+). Let the dropping point be the origin (0,0).
Initial velocity $u = +10\text{ m/s}$
Displacement $s = -40\text{ m}$ (since the ground is below the starting point)
Acceleration $a = -g = -9.8\text{ m/s}^2$

Step 2: Find the time taken ($t$) using $s = ut + \frac{1}{2}at^2$.

$$-40 = 10t + \frac{1}{2}(-9.8)t^2$$

$$-40 = 10t - 4.9t^2$$

Rearrange into standard quadratic form ($ax^2 + bx + c = 0$):

$$4.9t^2 - 10t - 40 = 0$$

Step 3: Solve the quadratic equation using the formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

$$t = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(4.9)(-40)}}{2(4.9)}$$

$$t = \frac{10 \pm \sqrt{100 + 784}}{9.8} = \frac{10 \pm \sqrt{884}}{9.8}$$

$$t = \frac{10 \pm 29.73}{9.8}$$

We ignore the negative root since time cannot be negative.

$$t = \frac{39.73}{9.8} \approx 4.05\text{ s}$$

$$\text{Time taken to reach ground } = 4.05\text{ s}$$

Step 4: Find the velocity upon impact ($v$).
We can use $v^2 = u^2 + 2as$ (which is less prone to rounding errors from $t$).

$$v^2 = (10)^2 + 2(-9.8)(-40)$$

$$v^2 = 100 + 784 = 884$$

$$v = \pm \sqrt{884} \approx \pm 29.73\text{ m/s}$$

Since the packet is moving downwards when it hits the ground, we take the negative sign.

$$\text{Impact Velocity } = 29.73\text{ m/s (downwards)}$$