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Step-by-Step Solution
Step 1: Determine the time and position of collision
Let the ground be our reference (i.e., height = 0). The piece of wood (mass 0.03 kg) is dropped from the top of a 100 m high building with an initial velocity of 0 m/s (downwards). The bullet (mass 0.02 kg) is fired upwards from the ground with an initial velocity of 100 m/s.
• Position of wood after time $t$: $y_{\text{wood}}(t) = 100 - \frac{1}{2} g t^2.$
• Position of bullet after time $t$: $y_{\text{bullet}}(t) = 0 + 100\,t - \frac{1}{2} g t^2.$
For collision, their positions must be equal:
$100 - \tfrac{1}{2} g t^2 = 100\,t - \tfrac{1}{2} g t^2 \quad \Rightarrow \quad 100 = 100\,t \quad \Rightarrow \quad t = 1\text{ s.}$
Hence, they collide 1 second after they start moving.
Step 2: Calculate the height at which the collision occurs
Using $y_{\text{wood}}(1)$ to find the collision point above the ground:
$y_{\text{wood}}(1) = 100 - \tfrac{1}{2} \times 10 \times 1^2 = 100 - 5 = 95 \text{ m.}$
So, the collision happens at 95 m above the ground, which is 5 m below the top of the building.
Step 3: Find the velocities of the wood and bullet just before collision
• Velocity of the wood after 1 s (downwards): $v_{\text{wood}} = 0 + g \times 1 = 10 \text{ m/s downward.}$
• Velocity of the bullet after 1 s (upwards): $v_{\text{bullet}} = 100 - g \times 1 = 100 - 10 = 90 \text{ m/s upward.}$
Step 4: Apply conservation of momentum at the instant of collision
Let upward direction be positive.
• Mass of wood $= m_w = 0.03\text{ kg}$ and its velocity just before collision $= -10 \text{ m/s}$ (negative, since downward).
• Mass of bullet $= m_b = 0.02\text{ kg}$ and its velocity just before collision $= +90 \text{ m/s}$.
Total momentum before collision:
$P_{\text{before}} = m_w \,(-10) + m_b \,(+90) \;=\; (0.03)(-10) + (0.02)(90) \;=\; -0.3 + 1.8 \;=\; 1.5 \text{ kg·m/s.}$
The combined mass immediately after collision is $m_w + m_b = 0.03 + 0.02 = 0.05\text{ kg}.$
Hence, velocity of the combined system immediately after collision:
$v_{\text{combined}} = \frac{P_{\text{before}}}{m_w + m_b} \;=\; \frac{1.5}{0.05} = 30 \text{ m/s (upwards).}$
Step 5: Calculate the additional height gained after collision
After collision, the combined system moves upwards with initial velocity 30 m/s. The additional height $h$ climbed before coming to rest is given by energy or kinematic relation:
$h = \frac{v_{\text{combined}}^2}{2\,g} = \frac{(30)^2}{2 \times 10} = \frac{900}{20} = 45 \text{ m.}$
Step 6: Find the maximum height above the top of the building
• The collision takes place 95 m above the ground, i.e., 5 m below the top of the building (top is at 100 m).
• The combined system rises an additional 45 m from the collision point.
• Therefore, the maximum height from the ground is $95 + 45 = 140 \text{ m.}$
• The building height is 100 m, so the rise above the building's top = $140 - 100 = 40 \text{ m.}$
Final Answer
The combined system rises 40 m above the top of the building.
