Question

Bookmar
A point mass of 1 kg collides elastically with a stationary point mass of 5 Kg. After the collision, the 1 Kg mass reverses its
direction and moves with a speed of 2 m/s. Which of the following statement is correct for the system of these two masses?
Total kinetic energy of the system is 6 J
O Momentum of the 5 Kg mass after collision is 4 Kg m/s
o Kinetic energy of the large mass after collision is 5 J
Total momentum of the system is 3 Kg m/s​

Answer 1

Answer:

Total momentum of the system is 3 Kg-m/s

Explanation:

If the velocity of 1 kg mass before collision is u and velocity of 5 kg mass after collision is v then

From the conservation of linear momentum

$1\times u+5\times 0=1\times (-2)+5\times v$

$\implies u=-2+5v$

Since the collision is elastic, total kinetic energy will be equal

Therefore,

$\frac{1}{2}\times 1\times u^2= \frac{1}{2}\times 1\times 2^2+\frac{1}{2}\times 5\times v^2$

$\implies u^2=4+5v^2$

$\implies (-2+5v)^2=4+5v^2$

$\implies 4-20v+25v^2=4+5v^2$

$\implies -20v+20v^2=0$

$\implies -20v(1-v)=0$

$\implies v=1$ m/s

Thus, $u=3$ m/s

The velocity of the centre of mass of combined system

$v_{cm}=\frac{m_1v_1+m_2v_2}{m_1+m_2}$

or, $v_{cm}=\frac{1\times (-2)+5\times 1}{1+5}$

$\implies v_{cm}=\frac{3}{6}$

$\implies v_{cm}=\frac{1}{2}$

The combined Kinetic Energy of the system

$KE=\frac{1}{2}(1+5)\times(\frac{1}{2}) ^2$

$KE=\frac{6}{8}$

$\implies KE=0.75$ J

Total momentum of the system

$=(m_1+m_2)v_{cm}$

$=(1+5)\times \frac{1}{2}$

$=3$ kg-m/s

Hope this helps.

Answer 2

Answer:

Explanation: hopefully....