Question

Block 1, of mass m1 = 2.70 kg , moves along a frictionless air track with speed v1 = 27.0 m/s . It collides with block 2, of mass m2 = 53.0 kg , which was initially at rest. The blocks stick together after the collision. A: Find the magnitude pi of the total initial momentum of the two-block system. Express your answer numerically. B: Find vf, the magnitude of the final velocity of the two-block system. Express your answer numerically. C: what is the change deltaK= Kfinal- K initial in the two block systems kinetic energy due to the collision ? Express your answer numerically in joules.

Answer 1

Answer:

a) Block 1 = 72.9kgm/s

Block 2 = 0kgm/s

b) vf = 1.31m/s

c) ∆KE = 936.36Joules

Explanation:

a) Momentum = mass× velocity

For block 1:

Momentum = 2.7×27

= 72.9kgm/s

For block 2:

Momentum = 53(0) (body is initially at rest)

= 0kgm/s

b) Using the law of conservation of momentum

m1u1+m2u2 = (m1+m2)v

m1 and m2 are the masses of the block

u1 and u2 are their initial velocity

v is the common velocity

Given m1 = 2.7kg, u1 = 27m/s, m2 = 53kg, u2 = 0m/s (body at rest)

2.7(27)+53(0) = (2.7+53)v

72.9 = 55.7v

V = 72.9/55.7

Vf = 1.31m/s

c) kinetic energy = 1/2mv²

Kinetic energy of block 1 = 1/2×2.7(27)²

= 984.15Joules

Kinetic energy of block 2 before collision = 0kgm/s

Total KE before collision = 984.15Joules

Kinetic energy after collision = 1/2(2.7+53)1.31²

= 1/2×55.7×1.31²

= 47.79Joules

∆KE = 984.15-47.79

∆KE = 936.36Joules