Block 1 of mass 0.200 kg sliding to the right over a frictionless elevated surfa
ID: 1523575 • Letter: B
Question
Block 1 of mass 0.200 kg sliding to the right over a frictionless elevated surface at a speed of 8.00 m/s. The block undergoes an elastic collision with stationary block 2, which is attached to a spring of spring constant 1208.5 N/m. (Assume that the spring does not affect the collision.) After the collision, block 2 oscillates in SHM with a period of 0.140 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 = 4.90 m. What is the value of d?
The mass of block 2 was solved to be 0.59 kg.
How can I solve for d using the equation v1+v1'=v2' 2/v2 (after mass cancels out)
Explanation / Answer
for SHM
T = 2/
and
= (k/m)
(k/m) = 2/T
k/m = 4² / T²
m = kT² / 4²
m = (1208.5)0.140² / 4²
m = 0.5999 kg
The center of mass (COM) before the collision and immediately after the collision until the spring force begins to affect the system is
v = 0.200(8.00) / (0.200 + 0.5999)
v = 2.00 m/s
The COM sees the 0.200 kg mass approaching at
8.00 - 2.00 = 6.00 m/s
as an elastic collision returns all the kinetic energy back to the mass which collided with the COM, the 0.200 kg mass appears to depart from the COM at v = -6.00 m/s
so a ground based observer sees the COM velocity plus the 0.200 kg mass relative to the COM or a velocity of
2.00 + (- 6.00) = -4.00 m/s
The 0.200 kg mass fall for how long?
h = ½gt²
t = (2h/g)
t = (2(4.90) / 9.81)
t = 0.999 s
so the horizontal distance d is
d = vt
d = 4.00(0.999)
d = 3.001 m