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Step-by-Step Solution
Step 1: Understand the Physical Situation
A bullet of mass 0.1 kg, traveling horizontally at 20 m/s, hits and sticks to a block of mass 1.9 kg that is resting at the edge of a 1 m high table. We want to find the kinetic energy of the combined mass (bullet+block) just before it hits the floor.
Step 2: Apply Conservation of Linear Momentum After Collision
Since the bullet embeds in the block, the collision is perfectly inelastic. Initially, only the bullet has horizontal momentum. The combined system moves together with a common velocity $v_{\text{after}}$ immediately after the collision.
Using conservation of momentum in the horizontal direction:
$m_{\text{bullet}} \times v_{\text{bullet}} = (m_{\text{block}} + m_{\text{bullet}}) \times v_{\text{after}}$
Substituting the values:
$0.1 \,\text{kg} \times 20 \,\text{m/s} = (1.9 \,\text{kg} + 0.1 \,\text{kg}) \times v_{\text{after}}$
Thus, $2 = 2.0 \times v_{\text{after}} \quad \Rightarrow \quad v_{\text{after}} = 1 \,\text{m/s}.$
Step 3: Analyze the Motion Off the Table
Once the block-bullet system leaves the table, it has a horizontal velocity of 1 m/s and is subject to gravitational acceleration vertically downward.
Step 4: Determine the Time to Fall and the Vertical Velocity
The height of fall is $1 \,\text{m}$. The time $t$ to fall from rest under gravity is given by the free-fall equation:
$H = \frac{1}{2} g t^2 \quad \Rightarrow \quad 1 = \frac{1}{2} \times 10 \times t^2 \quad \Rightarrow \quad t^2 = 0.2 \quad \Rightarrow \quad t \approx 0.447 \,\text{s}.$
The vertical velocity $v_{\text{vertical}}$ just before hitting the ground is:
$v_{\text{vertical}} = g \times t = 10 \times 0.447 \approx 4.47 \,\text{m/s}.$
Step 5: Determine the Resultant Velocity Just Before Impact
Horizontally, the velocity remains $1 \,\text{m/s}$ (neglecting air resistance). Vertically, it is $4.47 \,\text{m/s}$. Therefore, the resultant velocity $v_{\text{final}}$ is the vector combination of the horizontal and vertical velocities:
$v_{\text{final}} = \sqrt{(1)^2 + (4.47)^2} \approx \sqrt{1 + 19.98} \approx \sqrt{20.98} \approx 4.58 \,\text{m/s}.$
Step 6: Calculate the Kinetic Energy Just Before Impact
The total mass of the block-bullet system is $2.0 \,\text{kg}$. Kinetic energy (KE) is given by:
$KE = \frac{1}{2} \,(m_{\text{block+bullet}})\, (v_{\text{final}})^2 = \frac{1}{2} \times 2.0 \times (4.58)^2.$
Numerically, $KE \approx 1 \times 21 \approx 21 \,\text{J}.$
Final Answer
The kinetic energy of the combined mass just before it strikes the floor is approximately 21 J.
