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

Here we use the formula :

(m₁u₁) + (m₂u₂) = (m₁+m₂)V

(40 × 3.5) + (65 × 0) = (40 + 65)V

140 = 105V

V = 140 ÷ 105

V = 1.33 m/s

Therefore the speed at which the both move is 1.33 m/s

Answer 2

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

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

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

$\green{\underline \bold{Given :}}$

$\tt : \implies Mass \: of \: girl( m_{1}) = 40 \: kg$

$\tt : \implies Mass \: of \:friend( m_{2}) = 65 \: kg$

$\tt: \implies Speed \: of \: girl (v_{1})= 3.5 \: m/s$

$\red{\underline \bold{To \: Find :}}$

$\tt: \implies Speed \: of \: couple (v_{c})=?$

• Given Question

$\tt \circ \: Speed \: of \: friend(v_{2}) = 0 \: m/s$

$\bold{Using \: conservation \: of \: momentum}$

$\tt: \implies m_{1} v_{1} + m_{2}v_{2} = ( m_{1} + m_{2}) v_{c}$

$\tt: \implies 40 \times 3.5 + 65 \times 0 = (40 + 65) v_{c}$

$\tt: \implies 140 + 0 = 105 \times v_{c}$

$\tt: \implies v_{c} = \frac{140}{105}$

$\green{\tt: \implies v_{c} =1.34 \: m/s}$

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