Block A of mass m is connected to block B of mass 7.5 kg by a light inextensible rope passing over a frictionless pulley - NSC Physical Sciences - Question 2 - 2023 - Paper 1

The free-body diagram for block B includes two forces:

1. Weight ($W_B$) acting downward, given by: $W_B = m_Bg = 7.5 kg \cdot 9.8 m s^{-2} = 73.5 N$
2. Tension ($T$) acting upward.

Label these forces clearly in the diagram with appropriate arrows indicating their direction, and also label the mass $m_B = 7.5 kg$.

Newton's Second Law of Motion states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. In simple terms, the greater the force acting on an object, the greater its acceleration will be, provided its mass remains constant.

For block B: Using Newton's Second Law:
$F_{net_B} = m_B a$

Where:

• $F_{net_B} = W_B - T$
• $m_B = 7.5 kg$ and $a = 3.88 m s^{-2}$:

$W_B - T = m_B a$ $73.5 N - T = 7.5 kg \cdot 3.88 m s^{-2}$ $T = 73.5 N - 29.1 N = 44.4 N$

Now for block A: Using Newton's Second Law: $F_{net_A} = T - W_A$

• Let W_A = mg, where m is the unknown mass.

$T - mg = ma_A$ $44.4 N - mg = m \cdot a_A$

Setting $a_A = 3.88 m s^{-2}$: $44.4 N - mg = m \cdot 3.88 m s^{-2}$

From the information for block A, we can solve for m:
$44.4 = m(9.8 + 3.88)$ Solving gives: $m \approx 4.06 kg$

To determine the maximum height reached by block A after block B reaches the ground:

Assuming that block A moves upward until all kinetic energy of block B is transferred. Using energy conservation:

d= height above the ground + height initially suspended

Using the formula: $h = \frac{v^2}{2g}$

Where:

• $v = 3.41 m s^{-1}$ (initial velocity of A when B hits the ground)
• $g = 9.8 m s^{-2}$

Substituting the values: $h = \frac{(3.41 m s^{-1})^2}{2 \cdot (9.8 m s^{-2})} \approx 0.563 m$

Thus, the total height reached by block A: $H_{total} = 1.5 m + 0.563 m = 2.063 m \approx 2.06 m$