Question

Item 1 1 of 4 A block of wood connected to a horizontal spring is at rest on a frictionless floor. The mass of the block is 5kg and the spring constant is 125 N/m. A bullet of mass 50g is fired at the block with horizontal velocity of 25 m/s. The bullet gets embedded in the block after hitting it. The block (with the bullet embedded in it) undergoes simple harmonic oscillation as a result of this collision Part A What is the amplitude of the resulting oscillation? 150 AX ? m Submit Request Answer Part B What is the total mechanical energy of the block (with the bulletinside) as it undergoes simple harmonic oscillation? V AED ? J

Answer 1

9t is given in question The mass of the block m = 5 kg The mass of bullet m₂ = 50g spring constant k = 125 N/m 50x10 kg The velocity of bullet before collision of block uz = 25 m/s Before collision 4,20 Uz Let M be the combined mass of bullet and Block After collision and v be the velocity then by conservation of linear momentum mi hmma After collision M=mitma Total linear Total linear momentum = momentum before collision after collision -fp V miu, + m2 42 mov Mami + m2) 580+ 50x102x25 VE miuit M242 M 5+ 50x103 Ve 1.25 5:05 = 0.247 V= 0.25 mis This is the cafter aliniona o of embeded bullet and block

Now embeded block will start moving with this velocity and luill Starts S.Him. part A Now let A be the amplitude of S.Hum. of maximum displacement the total Kinetic energy will be converted in potential Now in the case energy of spring Hence Ź Mu? ² k je? ¥Mu? = K KA? S: je =A in the case se of maxim displacement mum A = Mv2 (mitm2) v2 K. k (5+0.05) (-25) 0.315625 125 125 A А 11 0.002525 А = 0.05 m or A = 5.0 cm Part B • Total mechanical Energy of block (bullet inside Maximum kinetic energy dengan kometve) - CS Scanned WTEE CamScanner Mv2

T. JI 늦 (mitmJv2 = (5.os) (oras) 11 2 - X 0.3156 2. 015 구 0016 J 다 TE CS Scanned with CamScanner