Answer:
a) Vx = 1088m/s
b) Vy = -162.93m/s
c) 5246745J
Explanation:
Mass of unbroken body = 23kg
Its velocity along +ve X-axis = 130m/s
Mass of first broken body, m1= 9.4kg
Its velocity along +ve X-axis = 130m/s
Nass of 2nd broken body, m2 = 6.1kg
Its velocity long-lived X - axis = -550m/s
Mass of 3rd broken body = ?
m3 = (23 - 9.4 - 6.1)kg
m3 = 7.5kg
Let velocity along the x-axis = Vx
Let the velocity along the x-axis = Vy
Applying law of conservation of momentum along x-axis
a) m1×0 + m2×(-550) + m3×(Vx) =M × 130
9.4 × 0 + 6.1× (-550) + 7.5(Vx) = 23 ×130
0 + (-5170) + 7.5Vx = 2990
2990 + 5170 = 7.5Vx
8160 = 7.5Vx
Vx = 8160/7.5
Vx = 1088m/s
b) Aplying conservation of momentum along the x-axis
(m1×130) + (m2 × 0) + (m3× Vy) = 0
(9.4 × 130) + (6.1 ×550) + 7.5Vy = 0
1222 + 0 + 7.5Vy = 0
1222 = -7.5Vy
Vy = 1222/(-7.5)
Vy = -262.93m/s
c) The energy released or change in KE is given by:
1/2[(m1v1^2) + (m2v2^2) +(m3Vx^2) ]= MV^2
Change in KE = 1/2[ 9.4× 130^2 + 6.1 × 550^2 + 7.5 × 1088^2 ] - 1/2(23 × 130^2)
Change in KE = 1/2[158860 + 1845250 + 8878080] - 1/2[388700]
Change in KE = 5441095 - 194350
Change in KE = 5246745J
