Answer:
yeah I'm Pretty sure it's b
Answer:
7.328m/s
Explanation:
Given parameters:
height of table = 0.68m
final velocity of the ball = 6m/s
Unknown:
Initial velocity of ball = ?
Solution:
To solve this problem, we are going to employ the appropriate motion equation.
We must understand that this fall occurs in the presence of gravity;
V = U + 2gH
Where;
V is the final velocity
U is the initial velocity
g is the acceleration due to gravity
H is the height of the pool table
Since U is the unknown, let us make it the subject of the expression;
U = V - 2gH
U = 6 - (2 x 9.8 x 0.68) = 7.328m/s(deceleration)
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
The second statement is the correct choice. Don't make me type it out.
