My Notes C Ask Your Teacher -5 points CJ9 7.P.045 Starting with an initial speed

Question

My Notes C Ask Your Teacher -5 points CJ9 7.P.045 Starting with an initial speed of 5.00 m/s at a height of 0.310 m a 1.55 kg ball swings downward and strikes a 4.75-kg ball that is at rest, as the drawing shows. 1 kg hm 5.00 m/s m2 kg a) Using the principle of conservation of mechanical energy, find the speed of the 1.55-kg ball just before mpact m/s (b) Assuming that the collision is elastic, find the velocities magnitude and direction of both balls just after the collision m/s (1.55 kg-ba m/s (4.75 kg-ball) (C) How high does each ball swing after the collision, ignoring air resistance m (1.55 kg-ball) m (4.75 kg-ball)

Explanation / Answer

a) Height of the ball 1(h)= 0.310 and mass m1 = 1.55 kg
Potential energy posses by the mass m1 = m1gh
this potential energy converted into kinetic energy
therefore (1/2)m1V2 = m1gh
V = (2gh)1/2 =2.466 m/s
(b) Since the collision is elastic therefore both momentum and energy remain conserve
so applying conservation of momentum
m1V1 + m2V2 = m1V1f + m2V2f
(1.55*5) +(4.75*0) = m1V1f + m2V2f
1.55V1f + 4.75 V2f = 7.75
V1f = 5 - 3.06 V2f
now applying Conservation of kintetic energy
(1/2)m1V12 + (1/2)m2V22 = (1/2)m1V1f2 + (1/2)m2V2f2
(1/2)(1.55)(25) = 0.775V1f2 + 2.375V2f2
19.375 = 0.375(5 - 3.06 V2f)2 +  2.375V2f2
On solving V2f = 2.6 m/s and V1f = -2.956 m/s
(c) we know that V = (2gh)1/2
so for ball of mass m1 , the height h1 = 0.445 m
for ball of mass m2 the height h2 = 0.344 m

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