Question

3. A block of mass m = 6.2 kg, moving on a frictionless surface with a velocity of -6.5 m/s to the right, collides with a block of mass M at rest, as shown in the figure. After the collision, the 6.2-Kg block recoils with a velocity of f =0.70 m/s to the left. If the blocks are in contact for 0.30 s, what is the magnitude of the average force on the 6.2-kg block, while the two blocks are in contact? M before after

Answer 1

Vi56.5 m/s [m] M m= 6.2 kq V20 f = -0.70m/s after before applying conservation Law of momentiun we eget, m ve + MV = mvp + MV Vi and if = initial and final relocity of mass m= 62 kg block the U=0 (for mass in block , before collision) = ? (for mass in black , after collision as w; and u are in opposite direction ,vo is taken negative ; f = 0.70 m/s

6.2 X 6.5 + OXM= 62x 60-70) + MV → MV = 6:2 (6:5+ 0,70) = 6,2x7' = 44.64 m.kg.5' Now for row from Newton and Law we know that force = rate of change of momentum = (MV) f - (MV): 44.64 - O as 0.30 slon \4 8 3 kg, m, s? - 148 g N impact of force x time = change of momentum 7 AP = Fit o = 44'64-0_ M -- Avg, magnitude of force applied repon M mass Hock by m=612 kg = 48.8 N from Newton 3rd Law, Mass M Hock , also and opposite give same amount of force revert back to block F = -48, 8N for m mass block but TIF/= 1488 NJ
​​​​​​