View Go Tools Window Help py 241 homework problems,160909.pdf (page 45 of 66) Co

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View Go Tools Window Help py 241 homework problems,160909.pdf (page 45 of 66) Collisions 1. Objects Thrown at a Pendulum (a) A pendulum initially hangs at rest below a support. Then a ball is thrown at the pendulum and it swings up to a final angle . Suppose we have a choice of throwing a "happy" ball or a "sad" ball at the pendulum. Assuming the balls have equal masses and are thrown with equal speeds, whichball will lead to a greater value of ? (Give a clear explanation) For parts (b) and (c), suppose an elastic ball and a clay ball have a mass of 200 g. The mass attached to the string is 1 kg. The balls are thrown at the hanging mass with a speed of 10 m/s. (b) Assume that the elastic ball undergoes an elastic collision with the hanging mass and the clay ball sticks to the hanging mass. Find the speed of the hanging mass immediately after the collision in both cases. (c) Assuming the string has a length of 1.25 m, find the greatest height which the hanging mass reaches, in each of the two cases. (The height is specified relative the initial height of the hanging mass at rest.)

Explanation / Answer

momentum of the ball is p = mv

the definition of happy ball and sad ball is not given,

in both cases the momentum is same.

Let us assume happy ball is perfectly elastic and sad ball is less elastic.

With the happy ball the momentum imparted to the pendulum is maximum and the sad ball being less leastic trnasfer less momentum to the pendulum, hence the angle is large in case of happy ball.

b) with elastic collision

vel. of the hanging mass immediately after collision

V2f   = 2m1/(m1+m2) V1i   = 2*0.2 /(1.2) *10

           = 3.33 m/s

inealstic collision with clay ball

intial momentum = 0.2 *10 = 2 kg-m/s

after collison the ball sticks to the mass and both move together

momentum after collision = 1.2 vf

conserving the momentum

1.2 vf = 2 ; vf = 2/1.2   = 1.67 m/s , vel. of hanging mass

c) elastic collison :

KE of the mass = 0.5*1* 3.332 = 5.54 J

let ha be max. heigth the ball reaches, conserving the energy

mgh = 5.54

h = 5.54/1*9.8 = 0.57 m

clay ball - inelastic collision

KE = 0.5 *1.2 *1.672 = 1.67 J

h = 1.67/ 1.2*9.8 = 0.14 m

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