Objects A and B are thrown vertically upwards with velocities v and 2v respectively. Object A reaches a maximum height h. What is the maximum height reached by object B? Ignore the effects of friction.

Step 1: Analyze Question 1.1. The gravitational force $F$ between two objects with masses $M$ and $m$, separated by a distance $r$, is given by Newton's Law of Universal Gravitation: $F = \frac{GMm}{r^2}$ where $G$ is the gravitational constant. For an object on the surface of a planet, $r$ is the radius of the planet. From the formula, $F$ is inversely proportional to $r^2$. Therefore, the gravitational force is inversely proportional to the square of the radius of the planet.

The correct option is C.

• C

Step 2: Analyze Question 1.2. The block starts from rest and moves to the right at constant acceleration. • If there is constant acceleration, the net force on the block is non-zero ($F_{net} = ma$, where $a > 0$). • Since the block is accelerating, its velocity is increasing. • Momentum is defined as $p = mv$. If velocity $v$ increases, then momentum $p$ increases. • Net work done $W_{net} = \Delta K = K_f - K_i$. Since velocity increases, kinetic energy $K = \frac{1}{2}mv^2$ increases, so $\Delta K > 0$. Thus, the net work done is not zero. • On a horizontal surface, the normal force is equal to the weight of the block, assuming no vertical acceleration.

Based on this analysis, the momentum of the block increases.

The correct option is B.

• B

Step 3: Analyze Question 1.3. For an object thrown vertically upwards, we can use the kinematic equation: $v_f^2 = v_i^2 + 2a\Delta y$ At maximum height, the final velocity $v_f = 0$. The acceleration $a = -g$ (due to gravity). The displacement $\Delta y$ is the maximum height $h_{max}$. Substituting these values: $0^2 = v_i^2 + 2(-g)h_{max}$ $0 = v_i^2 - 2gh_{max}$ $2gh_{max} = v_i^2$ $h_{max} = \frac{v_i^2}{2g}$ For object A: Initial velocity $v_A = v$. Maximum height $h_A = h$. So, $h = \frac{v^2}{2g}$ For object B: Initial velocity $v_B = 2v$. Let its maximum height be $h_B$. Using the formula for $h_{max}$: $h_B = \frac{(2v)^2}{2g}$ $h_B = \frac{4v^2}{2g}$ We can rewrite this as: $h_B = 4 \left(\frac{v^2}{2g}\right)$ Since $h = \frac{v^2}{2g}$, we substitute $h$ into the equation for $h_B$: $h_B = 4h$ The maximum height reached by object B is $4h$.

The correct option is D.

• D