A 0.454-kg block is attached to a horizontal spring that is at its equilibriumlength, and whose force constant is 24.0 N/m. The block rests on a frictionlesssurface. A 5.20x10-2-kg wad of putty is thrown horizontally at the block, hittingit with a speed of 8.96 m/s and sticking.Part AHow far does the putty-block system compress the spring?Templates Symbols undo redo reset keyboard shortcuts helpmax 14cm

Name: A 0.454-kg block is attached to a horizontal spring that is at its eq
Uploaded: 2024-03-20T06:57:25.000Z
Channel: Bartleby
Description: A 0.454-kg block is attached to a horizontal spring that is at its equilibriumlength, and whose force constant is 24.0 N/m. The block rests on a frictionlesssurface. A 5.20x10-2-kg wad of putty is thrown horizontally at the block, hittingit with a s

GivenMass of block, m1=0.454 kgBlock is at restInitial speed of block, u1=0Spring constant, k=24 N/mMass of wad of pulley, m2=5.2×10-2 kgInitial speed of wad of pulley, u2=8.96 m/sWhen there is collision between wad pulley and block they stick togetherLet v is the final velocity of wad pulley and block Momentum will remain conservedMomentum before collision=Momentum after collisionm1u1+m2u2=m1+m2vv=m1u1+m2u2m1+m2 =0.454×0+5.2×10-2×8.960.454+5.2×10-2v=0.9208 m/sPart ACalculate the distance pulley block system compresses the spring. xmaxBy the conservation of energyPotential energy of spring=Kinetic energy of pulley block system12kxmax2=12m1+m2v2kxmax2=m1+m2v2xmax2=m1+m2v2kxmax=m1+m2v2k =0.454+5.2×10-20.9208224 =0.1337 m =13.37 cm