Question

A 21.0kg block at rest on a horizontal frictionless table is connected to the wall via a spring with a spring constant k=34.0N/m. A 3.90×10-2kg bullet travelling with a speed of 550m/s embeds itself in the block. What is the amplitude of the resulting simple harmonic motion? Recall that the amplitude is the maximum displacement from equilibrium.

14. 1pt A 21.0kg block at rest on a horizontal frictionless table is connected to the wall via a spring with a spring constant k-34.0N/m. A 3.90x102kg bullet travelling with a speed of 550m/s embeds itself in the block. What is the amplitude of the resulting simple harmonic motion? Recall that the amplitude is the maximum displacement from equilibrium Answer: Submit All Answers Not yet correct, tries 1/6 Hint: Use conservation of momentum to get the initial speed of the bullet+block, then conservation of energy.

Answer 1

M = 21.0 kg u=0 k = 34.0 N/m mg = 3.90x102 kg ug = 550 m/s Use the law of conservation of momentum before the collision and after the collision of the bullet with the block as follows, myu; + Mu=(m, + M)v Rearrange the above equation for v and substitute the values in the equation. _mu: + Mu (m, +M) (3.90x102 kg (550 m/s)+(21.0 kg (0) (3.90x102 kg +21.0 kg =1.019 m/s Use the law of conservation of energy at the time of collision and after the collision as follows, KE, +PE: = KE, + PE, 3 (m, +M)v2 +0 = 0 + kk (m+ M )v? = kA Rearrange the above equation for A and substitute the values in the equation. (m. +Mv? (3.90x10-2kg +21.0 kg)(1.019 m/s)? (34.0 N/m) = 0.8015 m =80.15 cm Therefore, the amplitude of the resulting simple harmonic motion is 80.15 cm.