Calculate Speed of Clay Ball and 2kg Block After Collision

[AZhang]

A 2kg block is attached to an ideal spring (for which k = 200 N/m) and is initially at rest on a horizontal frictionless surface. A 100 gram ball of clay is thrown at the block. The clay is moving horizontally with speed V when it hits and sticks to the block. As a result, the pring is compressed a maximum distance of 0.4m.

a) Calculate the speed of the clay ball and 2kg block immediately after the clay sticks to the block but before the spring compresses significantly.


Ok, I kind of don't understand how the set-up is supposed to look like, so if someone can do ths problem, can they explain to me why they set it up the way they did?

THANKS :)

 

[vissh]

hello :D
The setup according to me is as -- http://s1102.photobucket.com/albums/g448/vissh/?action=view&current=temp.jpg"
See the pic and try to solve urself :D If by chance not able , see below [my answer might contain some mistake as i am too a learner :) ]

The box and spring are ideal as you said , so this will mean spring is uncompressed and unexpanded. Next, the clay is moving horizontly with velocity V (let). On collision,they stick with each other and move together thereafter [so i stick clay and box and let they acquired velocity v' as a new body of mass 2+0.1 = 2.1kg .]
By law of conservation of momentum,
2*0 + 0.1*V = 2.1*v'
=> V = 21*v'
The box+sticken clay as system
- Change in K.E. [from uncompressed to compressed spring]= 0 -(1/2)*(2.1)(v')²
............. =-(1/2)*(2.1)(v')²
- Work done by spring on system = -(1/2)kx²
........ = -(1/2)*200*(0.4)(0.4)
The net force on system is only of spring and thus By using work energy theorem (with the box+sticken clay as system):-
change in K.E. = Work done by spring
On solving you will get the answer :D

 

[AZhang]

I'm a little confused. Could you elaborate?