Hello i have these 2 question i need help with showing work . thank u 1) Mass m1

Question

Hello i have these 2 question i need help with showing work . thank u

1) Mass m1 = 2kg is initially moving with speed v1=0.90 m/s on a frictionless surface . Mass m2 = 1 kg is intially at rest. Mass m1 collides and sticks to mass m2 . They move together with velovity vf toward a spring of elastic constant K= 900 N/m

a- What is the speed of the two masses after collision ?
b-What is the magnitude of the change in momentum of mass m1 due to the collision with mass m2 ?
c-what is the change in kinetic energy of the two masses due to the collision ?
d-What is the maximum compression of the spring ?


2)- A block with mass m=22 kg rests on a frictionless table and is accelerated by a spring with spring constant K= 5420 N/m after being compressed a distance x1= 0.586 m from the spring's unstretched length .
The floor is frictionless except for a rough patch a distance d=3.6 m long . For this rough path , the coefficient of friction is uk=0.4 .
a-what is the speed of the block right after it leaves the spring ?
b-How much work done by friction as the block crosses the rough spot ?
c- what is the speed of the block after it passes the rough spot ?
d-what distance does the spring need to be compressed so that the block will just barely make it past the rough patch when released ?

Explanation / Answer

I can solve only 2nd question.

2) a) Work done by spring = 0.5 * 5420 * 0.586^2 = 930.6 J

Now 0.5 * 22 * v^2 = 930.6 =======> v = 9.2 m/s

b) Friction work = u * m * g * d = (0.4 * 22 * 9.8 * 3.6) = 310.5 J

c) The work done by the the friction force reduces the kinetic energy of the block.
Final KE = Initial KE - Work done by friction
0.5 * 22 * v^2 = (0.5 * 5420 * 0.586^2) - (0.4 * 22 * 9.8 * 3.6)
v^2 = [930.6 - 310.5] /11 = 56.4
v = 7.5 m/s

d) The work done by the spring must be "barely" greater than than the worjk done by the friction!
0.5 * k * x^2 = u * m* g * d
x^2 = (u * m* g * d) / (0.5 * k) = (0.4 * 22 * 9.8 * d) / (0.5 * 5420) = 0.0318*d
(here d is required to find x)

If the the spring is compressed barely more than x meter, the block will come to rest at a point barely past the d meter long rough patch!

If the rough patch is the original 3.6 m long:
x^2 = (u * m* g * d) / (0.5 * k) = (0.4 * 22 * 9.8 * 3.6) / (0.5 * 5420) =
x = 0.115 m

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