Question

Block 1 of mass 0.200g grams slides to the right with a speed of 8.00m/s. The block
undergoes an elastic collision with block 2, which is stationary and attached to a spring
with spring constant 1208.5 N/m. After the collision the block 2 oscillates with a period
of 0.140s and block 1 slides off the opposite end of the surface. If h = 4.90 m what is the
distance d that block 1 will go?

Answer 1

Use the period to find the mass of block 2
w = 2pi/T = 44.9 rad/s
w = sqrt(k/m)
m = k/w^2 = k*T^2 / (4pi^2) = (1208.5 N/m * (0.140s)^2) / (4*pi^2) = 0.60 kg

Use Conservation of Momentum + Elastic collision (KE is conserved) to find the velocity of block 1
m1*v1i + m2*v2i = m1*v1f + m2*v2f
m1*v1i^2 + m2*v2i^2 = m1*v1f^2 + m2*v2f^2

With v2i = 0
m1*v1i = m1*v1f + m2*v2f
m1*v1i^2 = m1*v1f^2 + m2*v2f^2

We have (2) equations with (2) unkowns (v1f and v2f).
Number crunching time
1.6 = 0.2*v1f + 0.6*v2f
12.8 = 0.2*v1f^2 + 0.6*v2f^2

Solve to top equation for v2f and sub it into the bottom equation
v2f = -0.2/0.6*v1f + 1.6/0.6 = -1/3*v1f + 8/3
12.8 = 0.2*v1f^2 + 0.6*(-1/3*v1f + 8/3)^2
12.8 = 0.2*v1f^2 + 0.6*(1/9*v1f^2 - 16/9*v1f + 64/9)
12.8 = 0.2*v1f^2 + 0.0667*v1f^2 - 1.0667*v1f + 4.2667
0 = (0.2+0.0667)*v1f^2 - 1.0667*v1f + (4.2667-12.8)
0 = 0.2667*v1f^2 - 1.0667*v1f - 8.5333

Use the quadratic equation for v1f
v1f = -4 m/s... NOTE: the other answer is 8 m/s but that is what it's initial velocity was which would imply a 0 final velocity for block 2. The negative implies it is in the opposite direction.

Now, since the surface is frictionless we know the horizontal component of the freefall velocity. Find the time it takes to fall.

Equation of motion in the vertical direction
hf = hi + vi*t - 1/2*gt^2

hf = 0 and vi (vertical component) = 0
0 = hi - 1/2*g*t^2
t = sqrt(2*hi/g) = sqrt(2 * 4.90 m / 9.81 m/s^2) = 1 s

Just use the time and horizontal velocity to find 'd'
d = v1f*t = 4 m/s * 1 s = 4 m