Suppose the following experiment is performed. A 0.250-kg object (m1) is slid on

Question

Suppose the following experiment is performed. A 0.250-kg object (m1) is slid on a frictionless surface into a dark room, where it strikes an initially stationary object with mass of 0.400 kg (m2) . The 0.250-kg object emerges from the room at an angle of 45.0o with its incoming direction. The speed of the 0.250-kg object is originally 2.00 m/s and is 1.50 m/s after the collision. We know that the magnitude of the direction of velocity of (v2 and 2) of the 0.400-kg object after the collision is 0.886 m/s and 312o. Calculate the internal kinetic energy of this two-object system before and after the collision. You will find that the internal kinetic energy is less after the collision, and so the collision is inelastic.

Explanation / Answer

let the object initially sliding towards +x axis

Apply conservation of momentum in y-direction

initial momentum = final momentum

0 = m1*v1fy + m2*v2fy

v2fy = -m1*v1fy/m2

= -0.25*1.5*sin(45)/0.4

= -0.663 m/s

apply conservation of momentum in x-direction

m1*v1i = m1*v1fx + m2*v2fx

0.25*2 = 0.25*1.5*cos(45) + 0.4*v2fx

v2fx = (0.25*2 - 0.25*1.5*cos(45))/0.4

= 0.587 m/s

v2f = sqrt(v2fx^2 + v2fy^2)

= sqrt(0.663^2 + 0.587^2)

= 0.886 m/s

KE_before = (1/2)*m1*v1^2

= (1/2)*0.25*2^2

= 0.5 J

KE_after = (1/2)*m1*v1f^2 + (1/2)*m2*v2f^2

= (1/2)*0.25*1.5^2 + (1/2)*0.4*0.886^2

= 0.438 J

clearly, KE_before > KE_after

so, the collision is inelastic.

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