A 28.0-kg body is moving in the direction of the positive x axis with a speed of
ID: 1777627 • Letter: A
Question
A 28.0-kg body is moving in the direction of the positive x axis with a speed of 360 m/s when, owing to an internal explosion, it breaks into three pieces. One part, whose mass is 9.5 kg, moves away from the point of explosion with a speed of 342 m/s along the positive y axis. A second fragment, whose mass is 3.0 kg, moves away from the point of explosion with a speed of 328 m/s along the negative x axis. What is the speed of the third fragment? Ignore effects due to gravity. How much energy was released in the explosion?
Explanation / Answer
Given
masses m = 28 kg , moving in the +ve x direction with velocity v = 360 m/s
after an internal explosion , m breaks in to three pieces say m1,m2,m3 moving in different directions with diffrerent velocities
from conservation of momentum the total momentum of the system before and after the explosion is same
momentum isa vector quantity so , the components of the momentum can be calculated
initial momentum before explosion is
P = Px + py
p = 28*360 + 0 = 10080 i + 0 j
now masses of pieces and their velcoicities are
m1 = 9.5 kg, v1 = 342 m/s along +y direction
m2 = 3.0 kg, v2 = 328 m/s along -x direction
m3 = (m-(m1+m2)) = 28-(9.5+3) = 15.5kg , v3 = ?
from conservation of momentum , for Vx3
Px = m1*vx1+m2*vx2+m3*vx3
10080 = m1*0 + 3(-328)+15.5(vx3)
Vx3 = 713.80 m/s
for Vy3
Py = m1*vy1+m2*vy2+m3*vy3
10080 = 9.5*342 + m2*0+15.5(vy3)
Vy3 = 440.70 m/s
the speed of third piece is v = sqrt(vx3^2+vy3^2) = sqrt(713.80^2+440.70^2) m/s = 838.88 m/s
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the initial kinetic energy of the 28 kg block is
k.e_i = 0.5*28*360^2 J = 1814400 J
final k.e , k.e_f = 0.5(m1*v1^2+m2*2^2+m3*v3^2)
= 0.5(9.5*342^2+3*328^2+15.5*838.88^2) J
= 6170782.322 J
the energy difference is
k.e_f - k.e_i = 6170782.322 -1814400 J = 4356382.322 J
the energy released in the explosion is 4356382.322 J
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