Question

A 7.3 kg block with a speed of 4.8 m/s collides with a 14.6 kg block that has a speed of 3.2 m/s in the same direction. After the collision, the 14.6 kg block is observed to be traveling in the original direction with a speed of 4.0 m/s. (a) What is the velocity of the 7.3 kg block immediately after the collision? (b) By how much does the total kinetic energy of the system of two blocks change because of the collision? (c) Suppose, instead, that the 14.6 kg block ends up with a speed of 6.4 m/s. What then is the change in the total kinetic energy

Answer 1

a.)

From momentum conservation,

Pi = Pf

m1*u1 + m2*u2 = m1*v1 + m2*v2

here given that, m1 = 7.3 kg

m2 = 14.6 kg

u1 = 4.8 m/s

u2 = 3.2 m/s

v2 = 4.0 m/s

v1 = ??

then, 7.3*4.8 + 14.6*3.2 = 7.3*v1 + 14.6*4.0

v1 = (7.3*4.8 + 14.6*3.2 - 14.6*4.0)/7.3

v1 = 3.2 m/s , in the original direction

b.)

Initial kinetic energy = Ki = 0.5*m1*u1^2 + 0.5*m2*u2^2

Ki = 0.5*7.3*4.8^2 + 0.5*14.6*3.2^2 = 158.848 J

Final kinetic energy = Kf = 0.5*m1*v1^2 + 0.5*m2*v2^2

Kf = 0.5*7.3*3.2^2 + 0.5*14.6*4.0^2 = 154.176 J

So, change in kinetic energy = Kf - Ki = 154.176 - 158.848 = -4.672 J

c.)

Now, given that v2 = 6.4 m/s

then, v1 = (m1*u1 + m2*u2 - m2*v2)/m1

v1 = (7.3*4.8 + 14.6*3.2 - 14.6*6.4)/7.3

v1 = -1.6 m/s , in the opposite direction of original motion

then, Final kinetic energy = Kf = 0.5*m1*v1^2 + 0.5*m2*v2^2

Kf = 0.5*7.3*(-1.6)^2 + 0.5*14.6*6.4^2 = 308.352 J

therefore, change in kinetic energy = Kf - Ki = 308.352 - 158.848 = 149.504 J

Since, in this case final kinetic energy is greater than initial kinetic energy.

So, speed of 6.4 m/s of the 14.6 kg block is not possible. (From energy conservation)

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