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Center of Mass and Collision question

2013 · Q144
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Center of Mass and Collision question

2013 · Q144

NEETPhysicsCenter of Mass and CollisionMCQ+4 / −1
An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass 1 kg moves with a speed of 12 m s−-−1 and the second part of mass 2 kg moves with 8 m s−-−1 speed. If the third part files off with 4 m s−-−1 speed, then its mass is :
  1. A
    7 kg
  2. B
    17 kg
  3. C
    3 kg
  4. D
    5 kg
View written solutionFree

Correct answer: D

The situation is as shown in the figure.

NEET 2013 Physics - Center of Mass and Collision Question 37 English Explanation

According to law of conservation of linear momentum

p→1+p→2+p→3=0{\overrightarrow p _1} + {\overrightarrow p _2} + {\overrightarrow p _3} = 0p​1​+p​2​+p​3​=0
∴\therefore∴ p→3=−(p→1+p→2){\overrightarrow p _3} = - \left( {{{\overrightarrow p }_1} + {{\overrightarrow p }_2}} \right)p​3​=−(p​1​+p​2​)

Here, p→1=(1 kg)(12 ms−1)i^=12i^ kg ms−1{\overrightarrow p _1} = (1\,kg)(12\,m{s^{ - 1}})\widehat i = 12\widehat i\,kg\,m{s^{ - 1}}p​1​=(1kg)(12ms−1)i=12ikgms−1

p→2=(2 kg)(8 ms−1)j^=16i^ kg ms−1{\overrightarrow p _2} = (2\,kg)(8\,m{s^{ - 1}})\widehat j = 16\widehat i\,kg\,m{s^{ - 1}}p​2​=(2kg)(8ms−1)j​=16ikgms−1

∴p→3=−(12i^+16j^) kgms−1 \therefore {\overrightarrow p _3} = - \left( {12\widehat i + 16\widehat j} \right)\,kgm{s^{ - 1}}∴p​3​=−(12i+16j​)kgms−1

The magnitude of p3 is

p3=(12)2+(16)2=20 kgms−1{p_3} = \sqrt {{{\left( {12} \right)}^2} + {{\left( {16} \right)}^2}} = 20\,kgm{s^{ - 1}}p3​=(12)2+(16)2​=20kgms−1

∴m3=p3v3=20 kgms−14ms−1=5kg \therefore {m_3} = {{{p_3}} \over {{v_3}}} = {{20\,kgm{s^{ - 1}}} \over {4m{s^{ - 1}}}} = 5kg∴m3​=v3​p3​​=4ms−120kgms−1​=5kg

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