| 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' = (2.0 kg • 3.5 m/s + 6.0 kg • -2.0 m/s)/8.0
kg |
|
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 g • 1 kg/1000 g = 0.01 kg m2
= 0.90 kg |
| v1
= 40.0 m/s
v2 = 0 |
| V'
= ?
V' = ? |
| |
| Δp
= 0 |
| pi
= pf |
| m1v1
+ m2v2 = m1v1' + m2v2'
= V'(m1 + m2) |
V' = (0.01 kg • 40.0 m/s)/8.01 kg
|
|
V' = 0.05 m/s |