An object with total mass m total = 16.5 kg is sitting at rest when it explodes
ID: 1363569 • Letter: A
Question
An object with total mass mtotal = 16.5 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.7 kg moves up and to the left at an angle of 1 = 19° above the –x axis with a speed of v1 = 26.6 m/s. A second piece with mass m2 = 5.2 kg moves down and to the right an angle of 2 = 24° to the right of the -y axis at a speed of v2 = 21.3 m/s.
1)
What is the magnitude of the final momentum of the system (all three pieces)?
2)
What is the mass of the third piece?
3)
What is the x-component of the velocity of the third piece?
4)
What is the y-component of the velocity of the third piece?
5)
What is the magnitude of the velocity of the center of mass of the pieces after the collision?
6)
Calculate the increase in kinetic energy of the pieces during the explosion.
Explanation / Answer
let,
masses, m1=4.7kg, m2=5.2 kg and m3
total mass is M=16.5 kg
M=m1+m2+m3
velocity is v1=26.6 m/sec, theta1=19 degrees,
v2=21.3 m/sec, theta2=24 degrees,
and
v3=(v3x)i+(v3y)j
1)
final momentum=iniial momentum =0
2)
total mass M=m1+m2+m3
16.5=4.7+5.2+m3
mass m3=6.6 kg
3)
mometum along horizontal line
m1*v1*cos(180-theta1)+m2*v2*cos(270+thet2)+m3*v3x=0
4.7*26.6*cos(180-19)+5.2*21.3*cos(270+24)+6.6*v3x=0
==> v3x=11.08 m/sec
4)
mometum along horizontal line
m1*v1*sin(180-theta1)+m2*v2*sin(270+thet2)+m3*v3y=0
4.7*26.6*sin(180-19)+5.2*21.3*sin(270+24)+6.6*v3y=0
==> v3y=9.16 m/sec
magnitude v3=sqrt(v3x^2+v3y^2)
V3=sqrt(11.08^2+9.16^2)
v3=14.37 m/sec
5)
here, velocity of C.M is , Vcm=0
6)
change in K.E dK=Kf- Ki
dK=(1/2*m1*v1^2+1/2*m2*v2^2+1/2*m3^2) - 0
=(1/2*4.7*26.6^2)+(1/2*5.2*21.3^2)+(1/2*6.6*14.37^2)
=3523.8 J