A block of mass m 1 = 2.9 kg initially moving to the right with a speed of 4.3 m

Question

A block of mass m1 = 2.9 kg initially moving to the right with a speed of 4.3 m/s on a frictionless, horizontal track collides with a spring attached to a second block of mass m2 = 5.2 kg initially moving to the left with a speed of 1.0m/s as shown in figure (a). The spring constant is 569 N/m.

(A) Find the velocities of the two blocks after the collision.
(B) During the collision, at the instant block 1 is moving to the right with a velocity of 0.7 m/s as in figure (b), determine the velocity of block 2.
(C) Determine the distance the spring is compressed at that instant.

(a) Find the maximum spring compression in this case.

b) What will be the individual velocities of the two masses (v1 and v2) after the spring extended fully again? (That is, when the two masses separate from each other after the collision is complete.)

The last part is all i need

(A) Find the velocities of the two blocks after the collision. Conceptualize: With the help of figure (a), run an animation of the collision in your mind. Figure (b) shows an instant during the collision when the spring is compressed. Eventually, block 1 and the spring will again separate, so the system will look like figure (a) again but with different velocity vectors for the two blocks Categorize: Because the spring force is conservative, kinetic en in the system is not transformed to internal energy during the compression of the spring. Ignoring any sound made when the block hits the spring, we can categorize the collision as being elastic. Analyze: Because momentum of the system is conserved, apply the equation: Substitute the known values: 2.9 kg) 4.3 m/s) (5.2 1.0 m/s) (2.9 kg)Vif (5.2 kg) v2f (1) 7.27 kg m/s 2.9 kg)Vif 5.2 kg)v2f Because the collision is elastic, apply the equation: (2) 4.3 m/s (-1.0 m/s) 5.3 m/s -virt var Substitute the known values: (3) 15.37 kg m/s 2.9 kg)vif (2.9 kg)v2f Multiply Equation (2) by 2.9 kg: 22.64 kg m/s J (8.1 kg) Add Equations (1) and (3): 22.64 kg m/s 2.80 m/s Solve for v2f 2f 8.1 kg Use Equation (2) to find vif: 5.3 m/s Vif 2.80 m/s v1f 2.5 m/s

Explanation / Answer

let's find the final velocities of both the masses after collision,

V1f = ( 2.9-5.2) 3.4 / (2.9+ 5.2) = -0.965 m/s apprx

v2f = 2( 2.9) ( 3.4)/ ( 8.1) = 2.434 m/s apprx

1/2 ( 2.9) ( 3.4)^2 = 1/ 2( 2.9) ( 0.965)^2 + 1/2 ( 5.2) ( 2.434)^2 + 1/2 (569) x^2

1/2 kx^2 = is apprxoxiamtely equal to 0 , which implies that there will be hardly any com,pression.

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