A horizontal spring of elastic constant k-60N/cm is compressed by 10 cm then rel

Question

A horizontal spring of elastic constant k-60N/cm is compressed by 10 cm then released. As it un-compresses the spring pushes on a block of mass m=0.5 kg overa frictionless horizontal table. The mass loses contact with the spring as the spring reaches its natural length. As it continues on its path the mass collides head- on and elastically with a second mass of 6kg initially at rest. If the 0.5 kg mass would bounce back, it would go towards the spring. If that is the case by how much will the mass compress the spring?

Explanation / Answer

Spring constatnt=60N/cm=6000N/m

compression in spring=10cm=0.1m

Initial energy stored in spring=0.5*6000*0.12=30 Joules

This is the enrgy imparted to the 0.5 kg block

Let velocity of block be v

0.5*0.5*v2=30

v=10.95 m/s

The speed of 0.5kg after elastic collision can be determined by momentum conseravtion and kinetic energy conservation

Initial momentum of system = 0.5*10.95=5.475 kg-m/s

Let final speeds of 0.5kg and 6 kg blocks be v1 and v2

From momentum conservation, 5.475=0.5v1+6v2

Or, 10.95 = v1+12v2 (i)

from kinetic energy conservation,

0.5*0.5*10.952=0.5*0.5*v12+0.5*6*v22

Or, 119.9025=v12+12v22 (ii)

From (i), v1=10.95-12v2

Substituting this in (ii),we get

119.9025=(10.95-12v2)2+12v22  

Or, 119.9025=119.9025-262.8v2+144v22+12v22

156v22-262.8v2 =0

v2(156v2-262.8)=0

v2=0 is not possible

v2=262.8/156=1.685 m/s

v1 = 10.95-(12*1.685)=-9.27 m/s

The 0.5kg block bounces back with a speed of 9.27 m/s

Let the compression of spring be x

0.5*6000*x2=0.5*0.5*9.272

x= 0.0846 m=8.46cm

Answer is 8.46 cm

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