So far I got that the apple was dropped 4.9m far down from the 12m top when the
ID: 1780656 • Letter: S
Question
So far I got that the apple was dropped 4.9m far down from the 12m top when the arrow hits it, and the velocity of the arrow when it hits the apple is 9.26m/s.. but i'm not too sure
please explain!
60. A 100-g apple is dropped from a height of 12 m and 1 second later is struck by a 100-g arrow flying upward at 15 m/s. (a) What is the speed of the apple and the arrow just after col- ow much energy is lost in the collision? (c) How long does it take the combined object to reach the ground from the time the apple was dropped ? ..Explanation / Answer
after falling 1 s
velocity of apple v1i = -gt j = -9.8*1j = -9.8 m/s j
velocity of arrow v2i = + 15 j m/s
maas of apple m1 = 100 g
mass of arrow = m2 = 100 g
after collision apple and srrow move together
from momentum conservation
momentum before collision = momentum after collision
m1*v1i + m2*v2i = (1m+m2)*vf
-(100*9.8 j)+ (100*15) j = (100+100)vf
vf = 2.6 j m/s
(a)
speed of apple and arrow = 2.6 m/s <<<<<=========ANSWER
(b)
initial kinetic energy Ki = (1/2)*m1*v1i^2 + (1/2)*m2*v2i^2
Ki = (1/2)*0.1*9.8^2 + (1/2)*0.1*15^2
Ki = 16.1 J
final kinetic energy Kf = (1/20*(m1+m2)*vf^2 = (1/2)*(0.1+0.1)*2.6^2 = 0.676 J
loss in energy = Ki - Kf = 15.424 J <<<<<=========ANSWER
--------------------
(c)
position of apple y -12 = -(1/2)*g*t^2 = -(1/2)*9.8*1^2 = -4.9 m
y = 12 - 4.9 = 7.1 m
after collision
along vertical
velocity voy = 2.6 m/s
displacement travelled by apple + arrow y = -7.1 m
acceleration ay = -g = -9.8 m/s^2
y = voy*t + (1/2)*ay*t^2
-7.1 = 2.6*t - (1/2)*9.8*t^2
t = 1.5 s
total time T = 1.5 + 1 = 2.5 s <<<<<=========ANSWER
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