The figure shows block 1 of mass 0.185 kg sliding to the right over a frictionle

Question

The figure shows block 1 of mass 0.185 kg sliding to the right over a frictionless elevated surface at a speed of 7.50 m/s. The block undergoes an elastic collision with stationary block 2, which is attached to a spring of spring constant 1130 N/m. (Assume that the spring does not affect the collision.) After the collision, block 2 oscillates in SHM with a period of 0.150 s, and block 1 slides off the opposite end of the elevated surface, landing a distance d from the base of that surface after falling height h = 6.40 m. What is the value of d?

Explanation / Answer

Use the period to find the mass of block 2
w = 2pi/T = 2x3.14/.150 s = 41.86 rad/s
w = sqrt(k/m)
m = k/w2 = k*T2 / (4pi2) = (1130 N/m * (0.150s)2) / (4*pi2) = 0.64 kg

Use Conservation of Momentum + Elastic collision (KE is conserved) to find the velocity of block 1
m1*v1 initial + m2* V2 initial = m1*v1final + m2*v2final
m1*v1 initial 2 + m2*V2 initial ^2 = m1*v1final2 + m2*v2final 2

With v2initial = 0
m1*v1 initial = m1*v1final + m2*v2final
m1*v1 initial2 = m1*v1final2 + m2*v2final 2

We have (2) equations with (2) unkowns (v1f and v2f).
Number crunching time

0.185kgx7.5m/s=1.39 = 0.185 kg*v1final + 0.64 kg *v2final =================>equation 1
0.185x7.52=10.40 = 0.185 kg *V1 final 2 + 0.64 kg *V2 final 2 ================>equation 2

Solve to top equation for v2final in equation 1 and substitute it into the equation 2
v2final = -0.185/0.64*v1final + 1.39/0.64 = -0.289*V1 final+ 2.171
10.40= 0.185*V1 final 2 + 0.64*(-0.289*V1 final+ 2.171)2
10.40 = 0.185*V1 final 2 + 0.64*(0.0835*V1 final2 - 1.254*V1 final+ 4.713)
10.40 = 0.185*V1 final 2 + 0.05344 V1 final2 - 0.80256*V1 final+ 3.016
0 = (0.185+0.05344 )*V1 final 2 - 0.80256*V1 final+ (3.016-10.40)
0 = 0.23844*V1 final 2 - 0.80256*V1 final - 7.384

Use the quadratic equation for V1 final
V1 final = -4.13 m/s... NOTE: the other answer is 7.49 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 * 6.40 m / 9.81 m/s^2) = 1.30 s

Just use the time and horizontal velocity to find 'd'
d = V1 final *t = 4.13 m/s * 1.30 s = 5.4 m

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