An object of mass m1 = 9.00 kg is in equilibrium while connected to a light spri

Question

An object of mass m1 = 9.00 kg is in equilibrium while connected to a light spring of constant k = 100 N/m that is fastened to a wall as shown in figure (a). A second object, m2 = 7.00 kg, is slowly pushed up against m1, compressing the spring by the amount A = 0.200 m, (see figure (b)). The system is then released, and both objects start moving to the right on the frictionless surface. 1. When m1 reaches the equilibrium point, m2 loses contact with m1 (see figure (c)) and moves to the right with speed v. Determine the value of v. 2. How far apart are the objects when the spring is fully stretched for the first time (D in figure (d))? This is non calculus based physics! Please don't solve with calculus

Explanation / Answer

1) conserving energy we get 0.5*k*A*A = 0.5*(m1+m2)*v*v

so v = 1/2 m/s

now let the speed of m1 be v1 and that of m2 be v2

momentum is conserved so (m1+m2)*v = m1*v1 + m2*v2

energy is conserved sp (m1+m2)*v*v = m1*v1*v1 + m2*v2*v2

solving we get v1 = v2 =1/2 m/s

so v = 1/2 m/s

2) when the spring is fully stretched conserving energy we get 0.5*9*v1*v1= 0.5*100*A*A

solving we get A =0.15 m

time taken by m1 to move 0.15 m = 0.2/(1/2) = 0.3 s

so distance covered by m2 = 0.3*1/2 = 0.15m

so the the distance between the blocks is 0 m

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