Ball A of mass 1 ~kg moving along a straight line with a velocity of 4 ~ms ⁻¹ hits another ball B of mass 3 ~kg which is at rest. After collision, they stick together and move with the same velocity along the same straight line. If the time of impact of the collision is 0.1 ~s then the force exerted on B is

Step-by-step solution

For ball A , aligned & m_A=1 ~kg , v_A=4 ~ms ⁻¹ & for ball B, m_B=3 ~kg , v_B=0 aligned Total momentum before collision, aligned p_i & =m_A v_A+m_B v_B & =1 4+3 0=4 ~kg - ms ⁻¹ aligned Since, after collision, both body stick together, therefore, it is the case of perfectly enelastic collision. Let v be the common velocity, then total momentum after collision, aligned p_f & = (m_A+m_B ) v & =(1+3) v=4 v aligned According to law of conservation of linear momentum, p_i=p_f 4=4 v v=1 ~ms ⁻¹ Time of impact, t=0.1 ~s For