Inelastic collision, spring compression

[peachpie]

A block of mass m1 = 1.8 kg slides along a frictionless table with a speed of 10 m/s. Directly in front of it, and moving in the same direction, is a block of mass m2 = 4.4 kg moving at 2.8 m/s. A massless spring with spring constant k = 1160 N/m is attached to the near side of m2, as shown in Fig. 10-35. When the blocks collide, what is the maximum compression of the spring? (Hint: At the moment of maximum compression of the spring, the two blocks move as one. Find the velocity by noting that the collision is completely inelastic to this point.)




velocity of center of mass = (m1v1 + m2v2)/ (m1 + m2), energy = 1/2mv^2 = 1/2kx^2



my velocity of center of mass = (1.8x10 + 4.4x2.8)/(1.8 + 4.4) = 4.89m/s, i plugged that into the energy equations [KE = 1/2(m1 + m2)(4.89)^2 = 1/2kx^2], and tried to solve for x. I also solved for x without finding the velocity of the center of mass using relative velocity; the faster object is traveling @ 7.2 m/s relative to the other object, so i plugged that into the energy equations. they were all wrong, and I'm not sure how to solve this.

 

[Doc Al]

my velocity of center of mass = (1.8x10 + 4.4x2.8)/(1.8 + 4.4) = 4.89m/s,

OK.

i plugged that into the energy equations [KE = 1/2(m1 + m2)(4.89)^2 = 1/2kx^2], and tried to solve for x.

No, you have it backwards. That's the amount of energy that did not go into compressing the spring. (Compare this to the initial KE of the system.)

Mechanical energy is conserved. At the point of maximum compression both masses move as one, but the system has both KE and spring potential energy.

 

[peachpie]

got it, thanks!
Could you also help me with this one?

Two 2.0 kg masses, A and B, collide. The velocities before the collision are vA = 20i + 25j and vB = -15i + 10.0j. After the collision, v'A = -3.0i + 18j. All speeds are given in meters per second.
How much kinetic energy was gained or lost in the collision?

I tried finding the total velocity of each particle (v = square root of vx^2 + vy^2) before and after the collision and plugging them into the energy equation, 1/2mv^2, then finding the difference of the before and after KE.

 

[Doc Al]

I tried finding the total velocity of each particle (v = square root of vx^2 + vy^2) before and after the collision and plugging them into the energy equation, 1/2mv^2, then finding the difference of the before and after KE.

That's fine, but first you have to find v'B. Did you?

 

[peachpie]

Yes, i found v'B to be <8i, 17j> (which was correct)

 

[Doc Al]

Yes, i found v'B to be <8i, 17j> (which was correct)

Then just tabulate the total KE before and after. Hint: You can calculate the KE directly from the components. KE = 1/2m(Vi^2 + Vj^2). That might save you some arithmetic and reduce the chance for errors.

 

[peachpie]

ahh, a calculation error...
i got it now!