Question

A 6.9 kg block with a speed of 3.6 m/s collides with a 13.8 kg block that has a speed of 2.4 m/s in the same direction. After the collision, the 13.8 kg block is observed to be traveling in the original direction with a speed of 3.0 m/s. (a) What is the velocity of the 6.9 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 13.8 kg block ends up with a speed of 4.8 m/s. What then is the change in the total kinetic energy

Answer 1

m1= 6.9 kg. m2= 13.8 kg

V1i= 3.6 m/s

V2i= 2.4 m/s. V2f= 3 m/s

a) According to the conservation of linear momentum,

(m1V1+m2V2)I=(m1V1+m2V2)f

1 and 2 are block 1 and block 2

V1f= (m1V1+m2V2)i-(m2V2)f/m1

=(6.9×3.6+13.8×2.4)-(13.8×3)/6.9

= 2.4 m/s

b) ∆K.E=(1/2 m1V1^2+1/2m2V2^2)f-.

( 1/2m1V1^2+1/2m2V2^2)i

∆K.E= (1/2×6.9×[3.6]^2+1/2×13.8×[2.4]^2)-(1/2×6.9×[3.6]^2+1/2×13.8×[2.4]^2

∆K.E=-2.484 J

c) V1f=(m1V1+m2V2)i- (m2V2)f/m1

=(6.9×3.6+13.8×2.4)-(13.8×4.8)/6.9

= -1.2 m/s

∆K.E=[ 1/2×6.9×(-1.2)^2+1/2×13.8×(4.8)^2]-[1/2×6.9×(3.6)^2+1/2×13.8×(2.4)^2]=69.552 J