Question

PLEASE SHOW WORK! A 4 kg body is traveling at 3 m/s with no external force acting on it. At a certain instant an internal explosion occurs, splitting the body into two chunks of 2 kg mass each. The explosion gives the chunks an additional 18 J of kinetic energy. Neither chunk leaves the line of original motion. Determine the speed and direction of motion of each of the chunks after the explosion. Hint: Translational momentum is conserved.

speed of slower chunk?
speed of faster chunk?

Answer 1

Step 1:

Using Momentum Conservation:

Pi = Pf

M*U = m1v1 + m2v2

M = mass of body = 4 kg

U = initial speed = 3 m/sec

m1 = m2 = 2 kg

v1 = final speed of m1

v2 = final speed of m2

So,

4*3 = 2*v1 + 2*v2

v1 + v2 = 6

Now Using Energy Conservation:

KE_final = KE_initial + KE_explosion

0.5*m1*v1^2 + 0.5*m2*v2^2 = 0.5*M*U^2 + KE_explosion

0.5*2*v1^2 + 0.5*2*v2^2 = 0.5*4*3^2 + 18

v1^2 + v2^2 = 36

Since we know v1 + v2 = 6

v2 = 6 - v1

v1^2 + (6 - v1)^2 = 36

v1^2 + 36 - 12*v1 + v1^2 = 36

2*v1^2 - 12*v1 = 0

2*v1*(v1 - 6) = 0

v1 = 6 & v1 = 0

So

Speed of slower chunk = 0 m/sec

Speed of faster chunk = 6 m/sec

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answered by: anonymous