Question

In the figure, block 2 of mass 2.60 kg oscillates on the end of a spring in SHM with a period of 26.00 ms. The position of the block is given by x = (1.50 cm) cos(ωt + π/2). Block 1 of mass 5.20 kg slides toward block 2 with a velocity of magnitude 3.00 m/s, directed along the spring's length. The two blocks undergo a completely inelastic collision at time t= 6.50 ms. (The duration of the collision is much less than the period of motion.) What is the amplitude of the SHM after the collision?

Answer 1

The spring conslant is, K= 45 mi - 4T (26) 72 (26x103)2 - 1S18X10 Non x= xmlos (7 + I)=-xm Y= 5.2(3)/(:675-a) = amis. K = 2 (7.8) (2) ² = 1567. 0-3 xx? (1518x105) (0-015) ? = 17:07 J- E=KFU= 156 +17.07 = 32'675. E = 1/2 ( 1.518 x105) a² 2 (32:67 1:518x105 x= 100am

Answer 2

n = 11-sum) Coss twit +74) diz - Casi) bow in /w* *74) at (t= 6.5mg) – 6:58163 26X103 .26X163. BS :72 .. ve - 11.5 kimi xare dips / 272 *6.5*10* 1 - 0 mis} lapply conservation of linear momentum) mix, tm, ve z'YM, 4 m) Vai 15-233 =15.242:6) Ven 15.6 :=-2 mis ) arte { Common_Velocity } at 4-6.5 ms (n = -1.5 cm)" lifune assume that duration of collision is much less than period of motion, then at t = 6.5 ms, we can assume that not velocity of combined mass I system is 2m/sec. E Apply Conservation of Energy Initial energy - Final energy E K x² + min. (k ) .

n where. (m'my ma) and K = (mwo?) K = 206 x (47%) < / 4X23.142?;,). :,(26)2 x 10-porty ( 26 x 10-493) · 21.516x10 (K = 1516) (m! - 7.8 kg) 15 16 X (1:5X264? + 7.8x4 = 1516 X, (x1)? (1-5X163)*+7.0*4 = (20)? i i 1516 = 2.25x154 + 205xióų. 720.7.25x164 = (? = 14.39 X 10² .=.nl . .. (1439) EX! 4.4 eur) new amphlude