The initial velocity component of block 2 is 2.50 m/s because the block is movin

Question

The initial velocity component of block 2 is 2.50 m/s because the block is moving to the left. The negative value for v2f means that block 2 is still moving to the left at the instant under consideration.

QUESTION Is it possible for both blocks to come to rest while the spring is being compressed? Explain. Hint: Look at the momentum in Equation (1). (Select all that apply.)

No, it is not possible.Yes, if the ratio of the two initial speeds of the blocks equals the inverse ratio of their masses.Yes, if the two blocks initially have equal masses.Yes, if the two blocks initially have initial momenta with equal magnitudes and opposite directions.Yes, if the two blocks initially have initial velocities with equal magnitudes and opposite directions.

PRACTICE IT

Use the worked example above to help you solve this problem. A block of mass m1 = 1.79 kg, initially moving to the right with a velocity of +4.32 m/s on a frictionless horizontal track, collides with a massless spring attached to a second block of mass m2 = 2.27 kg moving to the left with a velocity of

2.24 m/s,

EXERCISEHINTS:  GETTING STARTED  |  I'M STUCK!

Use the values from PRACTICE IT to help you work this exercise. Consider the instant that block 2 is at rest.

(a) Find the velocity of block 1. (Indicate the direction with the sign of your answer.)
v1f =  
What is the total momentum before the collision? Write an expression for the total momentum after the collision and solve for the velocity of block 1. m/s

(b) Find the compression of the spring.
x =  
Mechanical energy is conserved. Any kinetic energy that is missing from the system has been converted into potential energy in the spring. m

Explanation / Answer

(A) Applying momentum conservation,

(1.60 x 4) + (2.10 x - 2.50) = (1.60 x 3) + (2.10 v2f)

v2f = - 1.74 m/s

so 1.74 m/s to the left.

(B) Applying energy conservation,

PEi + KEi = PEf + KEf

0 + (1.60 x 4^2 / 2 + 2.10 x 2.50^2 / 2) = (1.60 x 3^2 /2 + 2.10 x 1.74^2 / 2) + ( 600 d^2 / 2)

d = 0.173 m

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Practice it: (using similar procedure)

(A) -1.20 m/s

(B) 0.213 m

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EXERCISE:

(A) (1.79 x 4.32) + (2.27 x -2.24) = (2.79 v) + 0

v = 0.95 m/s to the right

(B) x = 0.275 m

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