In the figure, block 2 (mass 1.20 kg) is at rest on a frictionless surface and t
ID: 2260831 • Letter: I
Question
In the figure, block 2 (mass 1.20 kg) is at rest on a frictionless surface and touching the end of an unstretched spring of spring constant 277 N/m. The other end of the spring is fixed to a wall. Block 1 (mass 1.60 kg), traveling at speed v1 = 4.70 m/s, collides with block 2, and the two blocks stick together. When the blocks momentarily stop, by what distance is the spring compressed?
In the figure, block 2 (mass 1.20 kg) is at rest on a frictionless surface and touching the end of an unstretched spring of spring constant 277 N/m. The other end of the spring is fixed to a wall. Block 1 (mass 1.60 kg), traveling at speed v1 = 4.70 m/s, collides with block 2, and the two blocks stick together. When the blocks momentarily stop, by what distance is the spring compressed?Explanation / Answer
mass of block1, m1 = 1.60 kg
mass of block2, m2 = 1.20 kg
initial velocity of block1, u1 = 4.70 m/s
initial velocity of block2, u2 = 0
final velocity of both after contact, v
by conservation of momentum
m1u1 + m2u2 = (m1 + m2)v
v = m1u1 / (m1 + m2) = 1.60*4.70 / (1.60 + 1.20) = 2.685 m/s
kinetic energy of blocks compresses the spring
K.E. of blocks = P.E. of spring
1/2 M(v^2) = 1/2 k(x^2) ...................k --- spring const; x distance compressed
or, x^2 = M(v^2) / k = 2.8(2.685^2) / 277 = 0.0728
or, x = sqrt(0.0728) = 0.2699 m = 26.99 cm