could you please solve qustion 3 ( i,ii,and iii) thank you (ii) Is the total lin

Question

could you please solve qustion 3 ( i,ii,and iii)
thank you

(ii) Is the total linear momentum of the two balls greater before or after the collision? (ii) Find the final velocities of the two balls after the collision. (iv) What is the velocity of the center of mass of the two balls before and after the collision? 3. (10 points) A pendulum made of a bob with mass 3 m" and a string of length "L" (assume this to be the distance from the pivot to the center of the bob) is released from a position where the string is horizontal as shown in the figure below. Upon reaching the bottom of its swing the pendulum bob strikes a sticky mass ..m , which was initially at rest, and sticks to it, (You may use m = 1 kg, and L to obtain numerical answers.) = 5 m (0) How high will the two mass combination pendulum swing before coming to a stop? (ii) Calculate the loss of kinetic energy in the collisicn (ii) Imagine that at the end of the swing (when the two masses are at height h), the sticky mass m falls off. Which of the following is true about the swing back of the pendulum? 1 of 7

Explanation / Answer

(i)At the highest point the bob has potential energy, which gets converetdt to kinetic energy at the bottom.

1/2 (3m) v^2 = (3m) g L

v = sqrt (2 g L) = sqrt (2 x 9.8 x 5) = 9.89 m/s

Intial KE of the system,

KEi = 1/2 3 m v^2 = 0.5 x 3 x 1 x 9.89^2 = 146.72 J

after the collision the stiky mass sticks to the bob and from conservation of momentum

3 m v = (3 m + m) vf

vf = 3 m v / 4 m = 7.42 m/s

KEf = 1/2 4 m vf^2

KEf = 0.5 x 4 x 1 x 7.42^2 = 110.11 J

The differnec in energgies is:

4m g h = 146.72 - 110.11 = 36.61 J

h = 36.61/4 x 9.81 = 0.934 m

Hence, h = 0.934 m

ii)The loss of kinetic energy is:

KE(loss) = 146.72 - 110.11 = 36.61 J

Hence, KE(loss) = 36.61 J

iii)from conservation of energy

1/2 3m v^2 = 4 m g h

So the velocity will increase

So, 4th option is correct.

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