A puck of mass m 1 = 0.151 kg slides along a frictionless horizontal surface wit

Question

A puck of mass m1 = 0.151 kg slides along a frictionless horizontal surface with a velocity of v1i = 0.450 m/s directed horizontally to the right. This puck makes a glancing collision with a second puck of mass m2 = 0.250 kg sliding with a velocity of v2i = 0.135 m/s directed horizonatlly to the left. After the collision, m1 is observed to be moving with a speed of v1f = 0.355 m/s at an angle of ø1f = 17.1 degrees counterclockwise from its original direction of motion (when viewed from above).

a) What is the velocity of puck m2 after the collision (both magnitude and direction)?

Explanation / Answer

here,

m1 = 0.151 kg

m2 = 0.25 kg

v1i = 0.45 m/s i

v2i = - 0.135 m/s i

v1f = 0.355 * ( cos(theta) i +sin(theta) j)

v1f = 0.34 i + 0.104 j m/s

a)

let the final speed of seccond puck be v2f

using conservation of momentum

m1 * v1i + m2 * v2i = m1 * v1f + m2 * v2f

0.151 * 0.45 i - 0.25 * 0.135 i = 0.151 * ( 0.34i + 0.104 j) + 0.25 * v2f

v2f = - 0.02 i - 0.06 j m/s

magnitude , |v2f| = sqrt(0.02^2 + 0.06^2)

|v2f| = 0.063 m/s

the angle , theta = arctan(0.06/0.02) = 252.3 degree

b)

initial kinetic energy , KEi = 0.5 * m1 * v1i^2 + 0.5 * m2 * v2i^2

KEi = 0.5 * ( 0.151 * 0.45^2 + 0.25 * 0.135^2) = 0.0175 J

KEf = 0.5 * ( 0.151 * 0.355^2 + 0.25 * 0.063^2) = 0.01 J

the kinetic energy is not conserved

so the collison is not elastic

the frcation change in kinetic energy , KEf= 0.57 * KEi

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