-- Option An object with total mass mtotal 17.1 kg is sitting at rest when it ex
ID: 2036748 • Letter: #
Question
-- Option An object with total mass mtotal 17.1 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 4.9 kg moves up and to the left at an angle of ?1-18. above the-x axis with a speed of v1 = 28.8 m/s. A second piece with mass m: 5.4 kg moves down and to the right an angle of ?2-23° to the right of the-y axis at a speed of v2-20.1 m/s. 1) What is the magnitude of the final momentum of the system (all three pieces)? kg-m/s Submit You currently have 0 submissions for this question. Only 9 submission are allowed. You can make 9 more submissions for this question. 2) What is the mass of the third piece? kg Submit You currently have 0 submissions for this question. Only 9 submission are allowed. You can make 9 more submissions for this question. 3) What is the x-component of the velocity of the third plece? m/s Submit You currently have 0 submissions for this question. Only 9 submission are allowed. You can make 9 more submissions for this question. 4) What is the y-component of the velocity of the third piece? m/s Submit You currently have 0 submissions for this question. Only 9 submission are allowed. You can make 9 more submissions for this question. 5) What is the magnitude of the velocity of the center of mass of the pieces after the collision? m/s Submit You currently have 0 submissions for this question. Only 9 submission are allowed. You can make 9 more submissions for this question. 6) Calculate the increase in kinetic energy of the pieces during the explosion SubmitExplanation / Answer
let
M = 17.1 kg
m1 = 4.9 kg, v1 = 28.8 m/s, theta1 = 18 degrees
m2 = 5.4 kg, v2 = 20.1 m/s, theta2 = 23 degrees
1) Final momentum of the system = initial momentum of the system before the collsion.
= 0
2) m3 = M - (m1+m2)
= 17.1 - (4.9 + 5.4)
= 6.8 kg
3) Apply conservation of momentum in x-direction
Pfx = Pix
-m1*v1*cos(theta1) + m2*v2*sin(theta2) + m3*v3x = 0
v3x = (m1*v1*cos(theta1) - m2*v2*sin(theta2))/m3
= (4.9*28.8*cos(18) - 5.4*20.1*sin(23))/6.8
= 13.5 m/s
4) Apply conservation of momentum in x-direction
Pfx = Pix
m1*v1*sin(theta1) - m2*v2*cos(theta2) + m3*v3y = 0
v3y = (-m1*v1*sin(theta1) + m2*v2*cos(theta2))/m3
= (-4.9*28.8*sin(18) + 5.4*20.1*cos(23))/6.8
= 8.28 m/s
5) v3 = sqrt(v3x^2 + v3y^2)
= sqrt(13.5^2 + 8.28^2)
= 15.8 m/s
6) increase in KE = KEf - KEi
= (1/2)*m1*v1^2 + (1/2)*m2*v2^2 + (1/2)*m3*v3^2 - 0
= (1/2)*4.9*28.8^2 + (1/2)*5.4*20.1^2 + (1/2)*6.8*15.8^2
= 3972 J
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