Block 1, of mass m 1 = 2.70 k g , moves along a frictionless air track with spee

Question

Block 1, of mass m1 = 2.70kg , moves along a frictionless air track with speed v1 = 31.0m/s . It collides with block 2, of mass m2 = 51.0kg , which was initially at rest. The blocks stick together after the collision.

Block 1, of mass m1 = 2.70kg , moves along a frictionless air track with speed v1 = 31.0m/s . It collides with block 2, of mass m2 = 51.0kg , which was initially at rest. The blocks stick together after the collision.

Explanation / Answer

a). total initial momentum = m1*v1 + m2*v2 = 2.7*31 + 51*0 = 83.7 kg-m/s b). from momentum conservation; 83.7 = (m1 + m2)*v => 83.7 = (2.7 +51)*v => v = 1.559 m/s c). initial K.E. = 0.5*m1*v1^2 = 0.5*2.7*31^2 = 1297.35 J final K.E. = 0.5*(2.7 +51)*(1.559)^2 = 65.258 J change in K.E. = 65.258 - 1297.35 = -1232.092 J

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