A block of mass m4 = 2.0 kg slides along a frictionless table with a speed of 10

Question

A block of mass m4 = 2.0 kg slides along a frictionless table with a speed of 10 m/s. Directly in front of it, and moving in the same direction, is a block of mass mB = 5.0 kg moving at 3.0 m/s. A massless spring with spring constant k = 1120 N/m is attached to the near side of mB, shown in Fig. 10-57. When the blocks collide, what is the maximum compression of the spring? (Hint: At the moment of maximum compression of the spring, the two blocks move as one. Find the velocity by noting that the collision is completely inelastic at this point.) FIGURE 10-57 = Problem 75.

Explanation / Answer

conserving momentum, Ma*Va + Mb*Vb = (Ma+Mb)*v so, v = (Ma*Va+Mb*Vb)/(Ma+Mb) = 5 m/s conserving energy, 0.5*Ma*Va^2 + 0.5*Mb*Vb^2 = 0.5*(Ma+Mb)*v^2 + 0.5*k*x^2 so, 0.5*2*10^2 + 0.5*5*3^2 = 0.5*(2+5)*5^2 + 0.5*1120*x^2 soving for x,we get, x = 0.25 m = 25 cm

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