A glider of mass 0.153 kg is moving to the right on a frictionless, horizontal a

Question

A glider of mass 0.153 kg is moving to the right on a frictionless, horizontal air track with a speed of 0.880 m/s . It has a head-on collision with a glider 0.300 kg that is moving to the left with a speed of 2.16 m/s . Suppose the collision is elastic. a)Find the magnitude of the final velocity of the 0.153 kg glider. b)Find the direction of the final velocity of the 0.153 kg glider. c)Find the magnitude of the final velocity of the 0.300 kg glider. d)Find the direction of the final velocity of the 0.300 kg glider.

Explanation / Answer

mass of the glider1 be M1 = 0.153 kg

mass of the glider2 be M2 = 0.3 kg

velocity of the glider1 be U1 = 0.88 m/s

velocity of the glider2 be U2 = - 2.16 m/s (it is moving opposite to M1)

since the collision is elastic both momentum and kinetic energy are conserved

Let V1 and V2 be velocities of gliders after collision

we can substitute this values in the below formula to get final velocity of glider1

V1 = (2 * M2 * U2)/(M1 + M2) +(( M1 - M2) * U1)/(M1 + M2)

= 2 * 0.3 * (-2.16)/(0.3 + 0.153) + (0.153 - 0.3)/(0.3 + 0.153)

= -286093 - 0.28556

= - 3.146 m/s

So Magnitude of final velocity 0.153 kg glider is 3.146 m/s

so the -ve sign suggest that first glider reverses its direction after collision i.e glider1 moves towards left after collision

final velocity of glider2

V2 = 2* M1 * U1/(M1+ M2) + ((M2 - M1) * U2)/(M1 + M2)

= (2 * 0.153 * 0.88)/(0.3 + 0.153) + (0.3 - 0.153) * (-2.16)/(0.3 + 0.153)

= 0.5944 - 0.70093

= - 0.106 m/s

magnitude of final velocity of second glider V2 = 0.106 m/s

-ve sign suggest that glider2 continues to move left after collision

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