Momentum Conservation: Practice Problem & Solution
A 6 kg box is travelling on ice at a speed of 9 m/s when a 12 kg packet is gently placed on it. The velocity will now be
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Momentum Conservation. Understanding the underlying formula is key to arriving at the correct answer below:
From Principal of conservation of momentum
m 1 v 1 = (m 1 + m 2 ) v
6 × 9 = ( 6 + 12) × v ⇒ v = 3 m/s
Detailed Solution (Best of Understanding Complex Concepts)
Step 1: Identify the given parameters
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Mass of the box ($m_1$) = 6 kg
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Initial velocity of the box ($u_1$) = 9 m/s
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Mass of the packet ($m_2$) = 12 kg
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Initial velocity of the packet ($u_2$) = 0 m/s (Because it is "gently placed," it has no initial horizontal velocity)
Step 2: Define the physics principle
Because the box is traveling on ice (implying no horizontal friction) and no external horizontal forces are acting on the system during the interaction, the total linear momentum of the system is conserved.
The formula for the conservation of linear momentum in this perfectly inelastic collision (since they move together) is:
$$P_{initial} = P_{final}$$
$$m_1u_1 + m_2u_2 = (m_1 + m_2)v$$
(where $v$ is the final common velocity of the box and the packet)
Step 3: Substitute the values into the equation
$$(6 \times 9) + (12 \times 0) = (6 + 12)v$$
$$54 + 0 = 18v$$
Step 4: Solve for the final velocity ($v$)
$$54 = 18v$$
$$v = \frac{54}{18}$$
$$v = 3 \text{ m/s}$$
The velocity of the combined box and packet will be 3 m/s.
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