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Work Energy and Power question

2015 · Q133
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Work Energy and Power question

2015 · Q133

NEETPhysicsWork Energy and PowerMCQ+4 / −1
Two particles A and B, move with constant velocities v1→\overrightarrow {{v_1}}v1​​ and v2→\overrightarrow {{v_2}}v2​​. At the initial moment their position vectors are r1→\overrightarrow {{r_1}}r1​​ and r2→\overrightarrow {{r_2}}r2​​ respectively. The condition for particles A and B for their collision is
  1. A
    r→1×v→1=r→2×v→2{\overrightarrow r _1} \times {\overrightarrow v _1} = {\overrightarrow r _2} \times {\overrightarrow v _2}r1​×v1​=r2​×v2​
  2. B
    r→1−r→2=v→1−v→2{\overrightarrow r _1} - {\overrightarrow r _2} = {\overrightarrow v _1} - {\overrightarrow v _2}r1​−r2​=v1​−v2​
  3. C
    r→1−r→2∣r→1−r→2∣=v→2−v→1∣v→2−v→1∣{{{{\overrightarrow r }_1} - {{\overrightarrow r }_2}} \over {\left| {{{\overrightarrow r }_1} - {{\overrightarrow r }_2}} \right|}} = {{{{\overrightarrow v }_2} - {{\overrightarrow v }_1}} \over {\left| {{{\overrightarrow v }_2} - {{\overrightarrow v }_1}} \right|}}​r1​−r2​​r1​−r2​​=​v2​−v1​​v2​−v1​​
  4. D
    r→1.v→1=r→2.v→2{\overrightarrow r _1}.{\overrightarrow v _1} = {\overrightarrow r _2}.{\overrightarrow v _2}r1​.v1​=r2​.v2​
View written solutionFree

Correct answer: C

For collision V→B/A{\overrightarrow V _{B/A}}VB/A​ should be along
B→A‾(r→A/B)\overline {B \to A} \left( {{{\overrightarrow r }_{A/B}}} \right)B→A(rA/B​)

So, V2→−V1→∣V2→−V1→∣=r1→−r2→∣r1→−r2→∣{{\overrightarrow {{V_2}} - \overrightarrow {{V_1}} } \over {\left| {\overrightarrow {{V_2}} - \overrightarrow {{V_1}} } \right|}} = {{\overrightarrow {{r_1}} - \overrightarrow {{r_2}} } \over {\left| {\overrightarrow {{r_1}} - \overrightarrow {{r_2}} } \right|}}​V2​​−V1​​​V2​​−V1​​​=∣r1​​−r2​​∣r1​​−r2​​​

AIPMT 2015 Physics - Work, Energy and Power Question 53 English Explanation

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