Question

Man A of mass 60kg pushes the other man B of mass 75lg due to which man B starts acceleration 3m/s².calculate the acceleration of man A at the instant.

Answer 1

accelration of man A is 3.75m/s^2

Answer 2

Given:

Mass of man A = 60 Kg

Mass of man B = 75 Kg

Velocity of man $B=3\frac { m }{ s }$

We Know that, according to Newton’s second law, $Force = Mass \times Acceleration$

Here external force is zero. According to law of conservation of momentum, total momentum before and after collision is equal.

Hence, $m_1 \times v_1 = m_2 \times v_2$

$60 \times v_1 = 75 \times 3$

$V_{ 2 }=75\times \frac { 3 }{ 60 } =3.75\frac { m }{ s }$