Question

A 40 kg girl skates at 3.5 m/s on ice toward her 65 kg friend who is standing still, with open arms. As they collide and hold each other, what is the speed of the couple?

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Speed\:of\:couple=1.34\:m/s}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt : \implies Mass \: of \: girl( m_{1}) = 40 \: kg \\ \\ \tt : \implies Mass \: of \:friend( m_{2}) = 65 \: kg \\ \\ \tt: \implies Speed \: of \: girl (v_{1})= 3.5 \: m/s \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Speed \: of \: couple (v_{c})=?$

• According to given question :

$\tt \circ \: Speed \: of \: friend(v_{2}) = 0 \: m/s\\ \\ \bold{Using \: conservation \: of \: momentum} \\ \tt: \implies m_{1} v_{1} + m_{2}v_{2} = ( m_{1} + m_{2}) v_{c} \\ \\ \tt: \implies 40 \times 3.5 + 65 \times 0 = (40 + 65) v_{c} \\ \\ \tt: \implies 140 + 0 = 105 \times v_{c} \\ \\ \tt: \implies v_{c} = \frac{140}{105} \\ \\ \green{\tt: \implies v_{c} =1.34 \: m/s} \\ \\ \green{\tt \therefore Speed \: of \: couple \: is \: 1.34 \: m/s}$