A 5.2 kg block with a speed of 15 m/s collides with a 15 kg block that has a spe

Question

A 5.2 kg block with a speed of 15 m/s collides with a 15 kg block that has a speed of 3.9 m/s in the same direction. After the collision, the 15 kg block is observed to be traveling in the original direction with a speed of 8.1 m/s. (a) What is the velocity of the 5.2 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 15 kg block ends up with a speed of 5.6 m/s. What then is the change in the total kinetic energy?

Explanation / Answer

let

m1 = 5.2 kg, m2 = 15 kg

before the collsion,
u1 = 15 m/s
u2 = 3.9 m/s

after the collsion,

v2 = 8.1 m/s

v1 = ?

a) Apply conservation of momentum

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

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

= (5.2*15 + 15*3.9 - 15*8.1)/5.2

= 2.88 m/s <<<<<<<-----------------------Answer

b) Ki = 0.5*m1*u1^2 + 0.5*m2*u2^2

= 0.5*5.2*15^2 + 0.5*15*3.9^2

= 699 J

Kf = 0.5*m1*v1^2 + 0.5*m2*v2^2

= 0.5*5.2*2.88^2 + 0.5*15*8.1^2

= 513.6

delta_KE = KI - Kf

= 699 - 513.6

= 185.4 <<<<<<<-----------------------Answer

c)

Apply conservation of momentum

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

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

= (5.2*15 + 15*3.9 - 15*5.6)/5.2

= 10.1 m/s <<<<<<<-----------------------Answer

b) Ki = 0.5*m1*u1^2 + 0.5*m2*u2^2

= 0.5*5.2*15^2 + 0.5*15*3.9^2

= 699 J

Kf = 0.5*m1*v1^2 + 0.5*m2*v2^2

= 0.5*5.2*10.1^2 + 0.5*15*5.6^2

= 500.4 J

delta_KE = KI - Kf

= 699 - 500.4

= 198.6 J <<<<<<<-----------------------Answer

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