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Standard XII

Physics

Relative Motion in 1D

82kg male and...

Question

$82 k g$ male and $48 k g$ female skate together at $7.4 m / s$ , male pushes female for a wall with $8.6 m / s$ Find the speed of male?

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Solution

Using the law of conservation of momentum :
Initial momentum = Final momentum or
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
$82 \times 7.4 + 48 \times 7.4 = 48 \times 8.6 + 82 \times v$
$(82 + 48) 7.4 = 48 \times 8.6 + 82 \times v$
$(130) 7.4 = 48 \times 8.6 + 82 v$
$v = 6.69 m / s$ .

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82kg male and 48kg female skate together at 7.4m/s, male pushes female for a wall with 8.6m/s Find the speed of male?

Using the law of conservation of momentum :Initial momentum = Final momentum orm1u1+m2u2=m1v1+m2v282×7.4+48×7.4=48×8.6+82×v(82+48)7.4=48×8.6+82×v(130)7.4=48×8.6+82vv=6.69m/s.

$82 k g$ male and $48 k g$ female skate together at $7.4 m / s$ , male pushes female for a wall with $8.6 m / s$ Find the speed of male?

Using the law of conservation of momentum :
Initial momentum = Final momentum or
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
$82 \times 7.4 + 48 \times 7.4 = 48 \times 8.6 + 82 \times v$
$(82 + 48) 7.4 = 48 \times 8.6 + 82 \times v$
$(130) 7.4 = 48 \times 8.6 + 82 v$
$v = 6.69 m / s$ .