A 5.1 kg block with a speed of 14 m/s collides with a 19 kg block that has a spe

Question

A 5.1 kg block with a speed of 14 m/s collides with a 19 kg block that has a speed of 6.2 m/s in the same direction. After the collision, the 5.1 kg block is observed to be traveling in the original direction with a speed of 8.3 m/s. (a) What is the velocity of the 19 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 5.1 kg block ends up with a speed of 7.3 m/s. What then is the change in the total kinetic energy?

Explanation / Answer

let

m1 = 5.1 kg, u1 = 14 m/s

m2 = 19 kg, u2 = 6.2 m/s

vafter the collison, v1 = 8.3 m/s

a) Apply conservation of momentum

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

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

= (5.1*14 + 19*6.2 - 5.1*8.3)/19

= 7.73 m/s

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

= 0.5*5.1*14^2 + 0.5*19*6.2^2

= 865 J

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

= 0.5*5.1*8.3^2 + 0.5*19*7.73^2

= 743 J

so, delta_KE = KEi - KEf

= 865 - 743

= 122 J <<<<<<<<-----------Answer

c)

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

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

= (5.1*14 + 19*6.2 - 5.1*7.3)/19

= 8 m/s

KEi = 0.5*m1*u1^2 + 0.5*m2*u2^2

= 0.5*5.1*14^2 + 0.5*19*6.2^2

= 865 J

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

= 0.5*5.1*7.3^2 + 0.5*19*8^2

= 743 J

so, delta_KE = KEi - KEf

= 865 - 743

= 122 J <<<<<<<<-----------Answer

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