A spring (k = 2900 N/m ) is compressed between two blocks: block 1 of inertia 1.

Question

A spring (k = 2900 N/m ) is compressed between two blocks: block 1 of inertia 1.80 kg and block 2 of inertia 2.00 kg. The combination is held together by a string (not shown in (Figure 1) ). The combination slides without spinning across low-friction ice at 2.90 m/s when suddenly the string breaks, allowing the spring to expand and the blocks to separate. Afterward, the 2.00-kg block is observed to move at a 34.0 angle to its initial line of motion at a speed of 3.50 m/s, while the smaller block moves off at an unknown speed and angle. Neither block is rotating after the separation, and you can ignore the inertias of the spring and the string relative to those of the blocks.

a)Determine the magnitude of velocity of block 1 after the separation.

b)Determine the direction of velocity of block 1 after the separation.

c)Determine the original compression of the spring, xx0, from its relaxed length.

Explanation / Answer

Given that

A spring with a consant (K) =2900N/m

block 1 of inertia (m1) =1.80 kg and block 2 of inertia (m2) = 2.00 kg

Now from the law of conservation of linear momentum

Initial momentum =Final momentum

(1.80kg+2.00kg)(2.90m/s)i =(2kg)(3.50)(cos34i+sin34j) +(1.80kg)v

now 11.02i=5.8032i+3.913j+1.80v

v =5.216i-3.913j/1.80 =2.898i-2.173j

a)

Now the magnitude of velocity of block 1 after separation is given by

v =Sqrt(2.898)2+(-2.173)2=Sqrt(8.398+4.725) =3.622m/s

b)

Determine the direction of velocity of block 1 after the separation is given by

Tantheta =vy/vx====>theta=tan-1(vy/vx) =tan-1(2.713/2.898) =43.11degrees is below -x -axis

c)

Now from the conservation of energy

(1/2)(m1+m2)vo2 =(1/2)k(x-xo)2 =(1/2)m1v12+(1/2)m2v22

Therefore the original compression of the spring, xx0, from its relaxed length is given by

x-xo =Sqrt(m1v12+m2v22-(m1+m2)vo2/k

=0.0799m =7.99cm

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