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...

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 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 <----------------------------------ANS