Answer:
(a) 96 ft/s
(b) - 3072 ft/s^2
(c) 0.03125 s
Explanation:
h = 144 ft
u = 0 ft/s
g = 32 ft/s^2
(a) let she strikes the box with velocity v.
Use third equation of motion
v = 96 ft/s
(b) Let the average acceleration is a.
initial velocity, u = 96 ft/s
final velocity, v = 0
h = 18 in = 1.5 ft
Use third equation of motion
a = - 3072 ft/s^2
(c) Let the time taken is t.
Use first equation of motion
v = u + at
0 = 96 - 3072 x t
t = 0.03125 second
Answer:
w_f = m*V*cos(Q_n) / L*(m+M)
Explanation:
Given:
- mass of the putty ball m
- mass of the rod M
- Velocity of the ball V
- Length of the rod L
- Angle the ball makes before colliding with rod Q_n
Find:
What is the angular speed ωf of the system immediately after the collision,
Solution:
- We can either use conservation of angular momentum or conservation of Energy. We will use Conservation of angular momentum of a system:
L_before = L_after
- Initially the rod is at rest, and ball is moving with the velocity V at angle Q from normal to the rod. We know that the component normal to the rod causes angular momentum. Hence,
L_before = L_ball = m*L*V*cos(Q_n)
- After colliding the ball sicks to the rod and both move together with angular speed w_f
L_after = (m+M)*L*v_f
Where, v_f = L*w_f
L_after = (m+M)*L^2 * w_f
- Now equate the two expression as per conservation of angular momentum:
m*L*V*cos(Q_n) = (m+M)*L^2 * w_f
w_f = m*V*cos(Q_n) / L*(m+M)
Answer:
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