Use equations of motion to get the correct answer!
3. The sum of the players' momenta is equal to the momentum of the players when they're stuck together:
(75 kg) (6 m/s) + (80 kg) (-4 m/s) = (75 kg + 80 kg) v
where v is the velocity of the combined players. Solve for v :
450 kg•m/s - 320 kg•m/s = (155 kg) v
v = (130 kg•m/s) / (155 kg)
v ≈ 0.84 m/s
4. The total momentum of the bowling balls prior to collision is conserved and is the same after their collision, so that
(6 kg) (5.1 m/s) + (4 kg) (-1.3 m/s) = (6 kg) (1.5 m/s) + (4 kg) v
where v is the new velocity of the 4-kg ball. Solve for v :
30.6 kg•m/s - 5.2 kg•m/s = 9 kg•m/s + (4 kg) v
v = (16.4 kg•m/s) / (4 kg)
v = 4.1 m/s
Answer:
The rate at which work is done is power
Answer:
<u>We are given:</u>
displacement (s) = 130 m
acceleration (a) = -5 m/s²
final velocity (v) = 0 m/s [the cars 'stops' in 130 m]
initial velocity (u) = u m/s
<u>Solving for initial velocity:</u>
From the third equation of motion:
v² - u² = 2as
replacing the variables
(0)² - (u)² = 2(-5)(130)
-u² = -1300
u² = 1300
u = √1300
u = 36 m/s