A 25.0-g object moving to the right at 23.0 cm/s overtakes and collides elastica

Question

A 25.0-g object moving to the right at 23.0 cm/s overtakes and collides elastically with a 10.0-g object moving in the same direction at 15.0 cm/s. Find the velocity of each object after the collision.


a 25.0-g object

cm/s



b 10.0-g object

cm/s

Explanation / Answer

Define variables: m1 = 26.0 g u1 = 25.0 cm/s , speed of m1 prior to collision v1 = ?, speed of m1 after collision m2 = 6.0 g u2 = 15.0 cm/s , speed of m2 prior to collision v2 = ?, speed of m2 after collision Need to perform a kinetic energy balance: Kinetic energy prior = Kinetic energy after ((m1*u1²)/2) + ((m2*u2²)/2) = ((m1*v1²)/2) + ((m2*v2²)/2) m1*u1² + m2*u2² = m1*v1² + m2*v2² m1*(u1² - v1²) = m2(v2² - u2²) m1*(u1 - v1)(u1 + v1) = m2(v2 - u2)(v2 + u2) =====> (i) Need to perform momentum balance: m1u1 + m2u2 = m1v1 + m2v2 m1(u1 - v1) = m2(v2 - u2) (v2 - u2) = m1(u1 - v1)/m2 =====> (ii) (u1 - v1) = m2(v2 - u2)/m1 =====> (iii) plug (ii) into (i) m1*(u1 - v1)(u1 + v1) = m2[m1(u1 - v1)/m2](v2 + u2) m1*(u1 - v1)(u1 + v1) = m2*m1(u1 - v1)*(v2 + u2)/m2 m1*(u1 + v1) = m1*(v2 + u2) solve for v2 m1*(u1 + v1) = m1*(v2 + u2) v2 + u2 = u1 + v1 v2 = u1 + v1 - u2 ====> (iv) plug (iv) back into momentum balance m1u1 + m2u2 = m1v1 + m2(u1 + v1 - u2) m1u1 + m2u2 -m2u1 +m2u2 = m1v1 + m2v1 v1(m1 + m2) = u1(m1-m2) + 2m2u2 v1 = (u1(m1-m2) + 2m2u2)/(m1+m2) v1 = (25*(26 - 6) + 2*6*15)/(26+6) = 21.25 cm/s Now use equation (iv) to find v2 v2 = u1 + v1 -u2 v2 = 25 + 15 - 21.25 = 18.75 cm/s sorry i used different values but formulaes are exactly same so that you can practice it yourself

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