Question

A skater of weight $30kg$ has initial speed $32m/s$ and second one of weight $40kg$ has $5m/s$. After the collision, they stick together and have a speed $5m/s$. Then the loss in KE is

• A. $48J$
• B. $96J$
• C. Zero
• D. None of these

Answer 1

The correct option is D None of these

Given:

Weight of the skater 1 is $m_1=30kg$

His initial speed $u_1=32m/s$

Weight of the skater 2 is $m_2=40kg$

His initial speed $u_2=5m/s$

After collision they stick together

Both of their speed $v_1=v_2=v=5m/s$

We know that kinetic energy $=\frac{1}{2}m{v}^2$

Initial kinetic energy $=K{E}_{skater1}+K{E}_{skater2}$

$=\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2$

$=\frac{1}{2}\left(30\right)(32{)}^2+\frac{1}{2}(40\left)\right(5{)}^2$

$=\frac{1}{2}\left(30\right)\left(1024\right)+\frac{1}{2}\left(40\right)\left(25\right)$

$=15360+500=15860J$

Final kinetic energy $=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2$

$=\frac{1}{2}{m}_1{v}^2+\frac{1}{2}{m}_2{v}^2=\frac{1}{2}(m_1+m_2)v^2$

$=\frac{1}{2}(30+40)(5{)}^2$

$=\frac{1}{2}\left(70\right)\left(25\right)=875J$

Loss in kinetic energy $\Delta KE=15860-875$

$=14985J$

The answer is none of the given options. So option D is the answer.