Two blocks are sliding on a horizontal frictionless surface with velocities show

Question

Two blocks are sliding on a horizontal frictionless surface with velocities shown in the sketch. Block A has mass 0.500 kg and block B has mass 0.400 kg. The two blocks have a perfectly inelastic collision and stick together after the collision.

(a) What is the speed of the combined blocks after the collision?


(b) What angle does the velocity of the combined blocks make with the +x-axis after the collision?


(c) What is the magnitude of the decrease in kinetic energy of the system of two blocks due to the collision?

Explanation / Answer

a) Use law of conservation of energy for perfectly inelastic collision,

P1i +P2i = P1f + P2f

m1v1i + m2v2i = m1v1f + m2v2f

For perfectly inelastic collision v1f = v2f = vcommon

m1v1i + m2v2i = (m1+m2)*vcommon

Along horizontal it becomes,

m1v1ix + m2v2ix = (m1+m2)*vcommonx

0.5*0.2 + 0.4*0 = (0.5+0.4)* vcommonx

vcommonx = 0.11 m/s

Along vertical it becomes,

m1v1iy + m2v2iy = (m1+m2)*vcommony

0.5*0 + 0.4*0.4 = (0.5+0.4)* vcommony

vcommony = 0.18m/s

vcommon = sqrt(vcommonx^2+ vcommony^2) = sqrt(0.11^2 + 0.18^2) = 0.21 m/s

b) = tan^-1(vcommony/vcommonx) = tan^-1(0.18/0.11) = 58.6 deg

c) KE = KEi – KEf = (1/2m1v1i^2 + 1/2m2v2i^2) – (1/2*(m1+m2)* vcommon2) = (1/2*0.5*0.2^2 + ½*0.4*0.4^2) – (1/2*(0.5+0.4)*0.21^2) = 0.022 J

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