1) A 2.0 kg block moving 3.5 m/s to the right collides with a 6.0 kg block moving 2.0 m/s to the left. Ignoring |
friction, determine the: |
(a) final velocity if the two blocks stick together. |
(b) amount of heat produced. |
(a) m1 = 2.0 kg m2 = 6.0 kg |
v1 = 3.5 m/s v2 = -2.0 m/s |
V' = ? V' = ? |
Δp = 0 |
pi = pf |
m1v1 + m2v2 = m1v1' + m2v2' = V'(m1 + m2) |
V' = ( |
V' = -0.62 m/s or 0.62 m/s to the left |
(b) ΔKE ≠ 0 |
KEi ≠ KEf |
1/2m1v2 + 1/2m2v22 ≠ 1/2m1v1'2 + 1/2m2v2'2 |
1/2•2.0 kg•(3.5 m/s)2 + 1/2•6.0 kg•(-2.0 m/s)2 ≠ 1/2•2.0 kg•(-0.62 m/s)2 + 1/2•6.0 kg•(-0.62 m/s)2 |
24.2 J ≠ 1.54 J |
ΔKE ≠ 0 |
ΔKE = 24.2 J - 1.54 J = 22.7 J of heat evolved |
You may be wondering why moving to the right is a positive velocity and to the left a negative velocity. |
There is no particular reason as long as you are consistent in your signs within the same problem. The |
same results would be obtained if you reversed the sign convention in the problem. Of course for the |
kinetic energy calculation it would not make any difference because the velocities are squared. |
2) A 10.0 g bullet is moving with a horizontal velocity of 40.0 m/s into a 8.0 kg block of wood which is at rest. If the |
bullet becomes embedded in the wood what is the final velocity of the wooden block? |
(a) m1 =
10.0 |
v1 = 40.0 m/s v2 = 0 |
V' = ? V' = ? |
Δp = 0 |
pi = pf |
m1v1 + m2v2 = m1v1' + m2v2' = V'(m1 + m2) |
V' = (0.01 |
V' = 0.05 m/s |