I need help with all parts and a force diagram if needed PROBLEM 2: A light ideal spring is attached to a block of mass m2 that sits at rest on an oily,frictionless table. A block of mass m, is then launched with an initial speed of v, at the springattached to block m2. Assuming no energy is lost due to friction, find(a) the speed of the blocks at the moment when the spring is maximally compressed, and(b) the fraction of m¡'s initial energy that is stored in the spring at that point, and(c) the speed of both blocks after m, detaches (bounces back) from the spring again.Part (2)Solve this numerically, based on your equations for the above problem. Given that m, is 3.77 kg, mjis 3.56kg, and vo is 4.7 m/s, find the final velocity (in m/s) of the spring block (m2) after the bounce as described inpart (c)Part (3)Is the collision in the problem above an elastic, inelastic, or perfectly elastic collision? How do you know?Explain your answer.

Hi there! Since there are multiple subparts in the question posted and it is not mentioned what is to be answered, we are providing the solution for the first 3 subparts. If the solution for a specified subpart is required please repost the question, specifying the particular subpart. Thank you!Given information: The mass attached to the spring = m2 The mass of the moving block = m1 The initial velocity of the moving block = v0 Part a: We know that, the moment when the spring is maximally compressed both the blocks will be in contact with the spring and both blocks will act as a single system and we know that the momentum is conserved in this case. So, the velocity of the system when the spring is compressed is given as follows: m1v0+m2(0)=(m1+m2)VV = m1(m1+m2)v0 The force diagram of the given scenario is: Since, given that the surface is oily, no energy lost due to friction, and the spring is ideal. We can say that the energy is conserved. So, we can write: KEm1=Spring energy + Kinetic energy of the system12m1v02=12kx2+12(m1+m2)V212kx2=12m1v02-12m12v02(m1+m2)=12m1v021-m1m1+m2spring energy =m2(m1+m2)KEm1 Let, the velocity of the "m1" after the collision is "v1" and "m2" be "v2", sinec the collision is considered as pure elastic we can say that the velocity of separation is equal to velocity of approach, we can write this as: v0-0=v2-v1⇒ v0=v2-v1 ........ Eq (1)And from the law of conservation of the momentum we can write:m1v0=m1v1+m2v2 ......... Eq (2)sunstututing Eq(1) in Eq(2) we get:m1v2-m1v1=m1v1+m2v2(m1-m2)v2=2m1v1v2=2m1v1(m1-m2)substituting this in equation 1 we get:v0=2m1v1m1-m2-v1⇒v0=v1m1+m2m1-m2v1=(m1-m2m1+m2)v0similarly v2 is given as:v2=(2m1m1+m2)v0

Answered: PROBLEM 2: A light ideal spring is…