, 10:40 PM * 97% Homework: Collisions, Impulse, And Reference Frames all Hits Wa

Question

, 10:40 PM * 97% Homework: Collisions, Impulse, And Reference Frames all Hits Wall 2 Optional 82 Anobject with 0etal mass moca 14.6issitting at rest when n explodes into three pieces. 0e piece with mass mt " 4.7 moves up and to the left at an angle of ,-20above the xaats with a speed of v-26,7m/s. Asecond piece with mass m-5.1 kg moves down and to the right an angle of -25 to the right of the y as at a speed of v 23.4 ) What is the magnitude of the final momentum of the system (all three pleces? 2) What is the mass of the third plece? What is he x-component of the velocity of the thind piecet What is the y-component of the velocity of the thind pliece 5) What is the magnibude of the velocity of the center of mass of the pieces after the collsion l Calculate the increase in kinetic energy of the pieces during the explosion

Explanation / Answer

Given that,

total mass of the object = M = 14.6 kg

m1 = 4.7 kg theta1 = 20 deg v1 = 26.7 m/s

m2 = 5.1 kg theta2 = 25 deg v2 = 23.4

Part(1)The magnitude of intial momentum is zero since the mass is at rest. And according to the conservation of linera momentum, the final momentum should equal to the intial. So the final momentum will also be zero.

Part (2) let m3 be the mass of third piece then,

m3 = M - (m1+m2) = 14.6 - (4.7 + 5.1) = 4.8 kg

Part(3)Let v3 be the x comp of velocity of m3

The momentum of m1 along x will be

Pxm1 = m1 v1 Cos theta1 = 4.7 x 26.7 x cos(20) = 117.92

and that of m2

Pxm2 = m2v2 sin(theta 2) = 5.1 x 23.4 x sin (25) = 50.44

So the x component of m3 will be :

Pxm3 = 117.92 - 50.44 = 67.48

So its velocity along x will be

v3x = Pxm3 / m3 = 67.48 /4.8 = 14.06 m/s

Part(4)Similar to part (3) the y component of v3 will be

m3V3y = m1v1sin(theta1) - m2v2cos(theta2) = 4.7 x 26.7 x sin(20) - 5.1 x 23.4 x cos25

m3V3y = 40.36 - 96.44 = -65.24 So

v3y = 65.24/4.8 = -13.59 m/s

Part(5) The momentum of system is zero, so the COM 's velocity V(com) = 0

Part(6)Increase in KE will be

v3 = sqrt (v3x^2 + v3y^2) = sqrt( 14.06^2 + -13.59^2) = 19.55m/s

KE = 1/2 [4.7 x 26.7^2 + 5.1 x 23.4^2 + 4.8 x 19.53^2) = 3.99 x 10^3 J

Hence, KE = 3.99 x 10^3 Joules

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