Question

A 0.10kg object moving initially with a velocity of 0.2m/s makes an elastic head on collision with a 0.15 kg object initially at rest. What percentage of the original

Answer 1

Answer: 4%

Explanation:

Initially, we have 0.1kg object moving with 0.2m/s and other one 0.15kg at rest.

Since, net external Force along horizontal direction on the system of these two objects is zero.

So, momentum along horizontal will be conserved.

We have,

$u_1=0.2m/s ,u_2=0\\ \\m_1=0.1kg ,m_2=0.15kg$

Now, As shown in figure.

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\\ \\=> 0.1*0.2+0.15*0=0.1*(-v_1)+0.15*v_2\\ \\=>2=-10v_1+15v_2$

2) Since, collision is elastic.

We know,

$e= \frac{v_{sep}}{v_{app}}\\ \\=>1=\frac{v_1+v_2}{0.2}\\ \\=>5v_1+5v_2=1$

3) Solving above two equations ,we get

$v_1=0.04m/s \\ \\v_2=0.16m/s$

4) We need to find percentage of original kinetic energy retained for 0.10kg object ,

That is,

$\frac{1/2mv_1^2}{1/2mu^2}*100=\frac{v_1^2}{u^2}*100\\ \\=\frac{0.04^2}{0.2^2}*100 \\ \\=4 \%$

Hence, our required retainment in kinetic energy of 0.10kg object is 4%.