Homework 5 Problems to hand in (show all your work! Examples: problems with mult

Question

Homework 5 Problems to hand in (show all your work! Examples: problems with multiple forces must show the FBDs! Problems with energies must show the conservation of energy or work-energy setup. Problems with momentum must show the conservation of momentum setup): CP1 Balls A (mass-4.0kg) and B (mass-1.0kg) both hang from chords R-1.5 meters in length. Ball B is initially at rest. Ball A is displaced at a 25 degree angle from vertical, and then, starting from rest, it is let go. What will be the maximum horizontal displacement from the bottom of the swing for each ball if a) the collision is head-on elastic? b) the collision is perfectly inelastic? What will be the maximum horizontal displacement from the bottom of the swing for each bal if R-1.5m (hint, consider conservation of energy and momentum)

Explanation / Answer

initially ball A at an angle of 25 deg with vertical.

height of ball A from ballB (lowest position)

h = R(1 -cos25)

using energy conservation to find speed of A at bottom (Or just before collision)

mgh = mv^2 /2

vA =sqrt(2 x 9.81 x 1.5(1-cos25)) = 1.66 m/s

a) If collision is elastic, then after collision suppose ball b goes with speed vB and ball A goes with vA.

in elastic colllision,

velocity of separation = velocity of approach

1.66 = vB - vA

vB = 1.66 + vA .......(i)

using momentum conservation,

4x1.66 + 1 x0 = 4vA + 1vB

putting vB from (i)

4x1.66 = 4vA + (1.66 + vA)

vA = 0.996 m/s

so vB = 1.66 + 0.996 = 2.60 m/s

for height using energy conservation,

mgh = mv^2 /2

and h = R(1 - cos@)

For A:

9.81 x 1.5 x (1 -cos@) = 0.996^2 / 2

@ = 14.92 deg

Horizontal displ. = Rsin@ = 1.5sin14.92 = 0.386 m ...........Ans for ball A

For B:

9.81 x 1.5 x (1 -cos@) = 2.60^2 / 2

@ = 39.62 deg

Horizontal displ. = Rsin@ = 1.5sin39.62= 0.957 m ...........Ans for ball B

b) for inelastic , now both ball will stick and will go together,

using momentum conservation,

4 x 1.66 = (4 +1)v

v = 1.328 m/s

mgh = mv^2 /2

and h = R(1 - cos@)

9.81 x 1.50 (1 - cos@) = 1.328^2 /2

@ = 19.94 deg

Horizontal displ. = Rsin@ = 1.5sin9.94=0.511 m

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