Question #192506

A 0.3-kg block is moving to the right on a horizontal, frictionless surface with a speed of 0.6 m/s. It makes a head-on collision with a 0.2-kg block that is moving to the left with a speed of 1.2 m/s. Find the final velocity (magnitude and direction) of each block if the collision is elastic. (Since the collision is head-on, all motion is along a line.)


1
Expert's answer
2021-05-14T09:42:58-0400

For an elastic collision


v1=m1m2m1+m2u1+2m2m1+m2u2=v_1=\frac{m_1-m_2}{m_1+m_2}\cdot u_1+\frac{2m_2}{m_1+m_2}\cdot u_2=


=0.30.20.3+0.20.6+20.20.3+0.2(1.2)=0.84 (m/s)=\frac{0.3-0.2}{0.3+0.2}\cdot 0.6+\frac{2\cdot0.2}{0.3+0.2}\cdot (-1.2)=-0.84\ (m/s) to the left


v2=m2m1m1+m2u2+2m1m1+m2u1=v_2=\frac{m_2-m_1}{m_1+m_2}\cdot u_2+\frac{2m_1}{m_1+m_2}\cdot u_1=


=0.20.30.3+0.2(1.2)+20.30.3+0.20.6=0.96 (m/s)=\frac{0.2-0.3}{0.3+0.2}\cdot (-1.2)+\frac{2\cdot0.3}{0.3+0.2}\cdot 0.6=0.96\ (m/s) to the right







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