If a skater of weight $3 \,kg$ has initial speed $32\, m/s$ and second one of weight $4 \,kg$ has $5 \,m/s$. After collision, they have speed (couple) $5\, m/s$. Then the loss in K.E. is
If a skater of weight $3 \,kg$ has initial speed $32\, m/s$ and second one of weight $4 \,kg$ has $5 \,m/s$. After collision, they have speed (couple) $5\, m/s$. Then the loss in K.E. is
$48 \,J$
$96\, J$
$Zero$
None of these
If a skater of weight $3 \,kg$ has initial speed $32\, m/s$ and second one of weight $4 \,kg$ has $5 \,m/s$. After collision, they have speed (couple) $5\, m/s$. Then the loss in K.E. is
Loss in K.E. = (initial K.E. -Final K.E.) of system
$\frac{1}{2}{m_1}u_1^2 + \frac{1}{2}{m_2}u_2^2 - \frac{1}{2}({m_1} + {m_2}){V^2}$
$ = \frac{1}{2}3 \times {(32)^2} + \frac{1}{2} \times 4 \times {(5)^2} - \frac{1}{2} \times (3 + 4) \times {(5)^2}$
$= 986.5 \,J$