Question

3) A 4.0-kg block starts from rest and slides down an incline, which make a 40'angle with the horizon, climbs up an incline that also makes a 40°angle with the horizon, and elastically collides with a 2.0-kg block. (See the figure.) How far apart do the blocks land on the horizon? (Assume there is no friction in the system.) 4.0 kg 2.0kg 10m 4.Om

Answer 1

13 h=10m/ .4.0m, mass M2 is m = 4.0kg. M₂= 2.0kg. Inclination, d= 40 Initial height of m, is, h = lom. Initially it is at rest, Final when heigh hit the ha = 4.om. If, when it hits m2 then - the relocity migh, maghz + I miu? m² = 29(hinha) u,ar zg Chicha) 2x9-8*(10-1), a, a 10.84 m/s Initial relocity ma is het I, and final relocity of m, and me elastie collision, ube be just aftes Se m,ut mal = m, et ma ₂, 4 x 10.84. On / 44, +20₂ = 21.68 2ut I Ising energy conservation 2, 2 & ma uz & molen Lee mu2 = + 4x(10-84) 40 & 282 f

using it Fromliy 20,= 21-68 a (21.68-V)2 to 262 4 x (10-84)? (21-68) 2x 21.68x4 + 362 = 4x (10.89 on, 3 V 43.36 V2 O, Ir 43.36 ཙུན 3 mi q= 14-45 m/s. when , is fully reflected back. Now , 21.68 - I 2 on, 3.61 m/s

to het, generalize the speeds So, restical component, as vocale). Pero Inz 98 dy Isino. B So, level, A Time to seach the same 2 usino 219 fia 9 g. Bu ya B ten from A is Time to seach at ground So, hes y ke + 1/2 gta (as at point A , restical velocity is same as e in dommmards) 2h2 = 0, ON, + 2 by to 88? on, -20ly t 4 g +8ghz, dz 2g lyt 92 +2ghz so, Total to hit ground, time taken T = titta. 2 103 -lgt T = + Gr + 2g ha g g ly ,ghܐ܀ ܨܶܟ + + of g. Horizontal distance drarelled is D = IT 19 cos 9x Т. vooso [ Usino + esino & rezsino + 2ghz fon 91 the distance of mi is -

D 3.6 1x costo TY [3:01 361 sin 40+ (3.61) 2 sim² 40 + 2x9.8x4. 9.8 on? Di 3-24 m. Similarly for me using ly D2= 14.45 x cos qol [14 14-45* Sin 40' + $(14.4 5 * singo)? +2x9.8*43] Dz 24.99m. 11 So Distance between the hitting point of both at ground ist AD – D2 - D. (24.99 - 3-24)m. masses AD 21.75 m