A(n) 7.8-kg object is sliding across the ice at 2.34 m/s in the positive x direc

Question

A(n) 7.8-kg object is sliding across the ice at 2.34 m/s in the positive x direction. An internal explosion occurs, splitting the object into two equal chunks and adding 16 J of kinetic energy to system. The explosive separation takes place over a 0.16-s time interval. Assume that the one of the chunks after explosion moves in the positive x direction. The x component of the average acceleration of this chunk during the explosion is a_front_x, the x component fo the average acceleration of the other chunk during the explosion is a_rear_x.

What are the x components of the average accelerations of the two chunks during the explosion?

Explanation / Answer

Mass of the object before the explosion,M =7.8 kg
Mass of the two pieces after explosion be m1 = m2 = M/2 = 3.9 Kg

Using Momentum Conservation -
Initial Momentum = M*v
Final Mometum = m1*v1 + m2 *v2

M*v = m1*v1 + m2 *v2
7.8 * 2.34 = 3.9 * v1 + 3.9 * v2
v1 + v2 = 4.68 -------------1

Intial Kinetic Energy = 0.5 * M*v^2
Final Kinetic Energy = 0.5 * m1v1^2 + 0.5 m2v2^2

Intial Kinetic Energy + 16 J = Final Kinetic Energy
0.5 * M*v^2 + 16 = 0.5 * m1v1^2 + 0.5 m2v2^2
0.5 * 7.8 * 2.34^2 + 16 = 0.5*3.9 * v1^2 + 0.5 * 3.9 * v2^2
v1^2 + v2^2 = 19.15

Substituting Value from eq 1 -
(4.68 - v2)^2 + v2^2 = 19.15
Solving for v2
v2 = 4.36 m/s
v1 = 0.32 m/s

Average Acceleration of Chunk 1 =
vfi = vin + a*t
4.36 = 2.34 + a*0.16
a = 12.625 m/s^2

Average Acceleration of Chunk 2 =
vfi = vin + a*t
0.32 = 2.34 + a*0.16
a = - 12.625 m/s^2


a_front_x = 12.625 m/s^2
a_rear_x = -12.625 m/s^2 (-ve sign means, towards -ve x axis.)

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