a. To calculate the speed of mass m just before it strikes mass M, we can first find the gravitational potential energy of mass m when it is at a height 0.30 m above the equilibrium position. This is given by:
PE_gravity = m * g * h
where m = 0.200 kg, g = 9.81 m/s², and h = 0.30 m.
PE_gravity = 0.200 * 9.81 * 0.30 = 0.5886 J
Since the ramp is frictionless, all of this potential energy gets converted into kinetic energy of mass m just before it strikes mass M. The kinetic energy of mass m at this point can be written as:
KE_m = 0.5 * m * v^2
where v is the speed of mass m just before it strikes mass M. From conservation of energy, we can write:
PE_gravity = KE_m
Substituting the expressions for PE_gravity and KE_m, we get:
0.5886 = 0.5 * 0.200 * v^2
Now we can solve this equation for v:
v^2 = (0.5886 / (0.5 * 0.200))
v^2 = 5.886
v = √5.886
v ≈ 2.426 m/s
So the speed of mass m just before it strikes mass M is approximately 2.426 m/s.
b. Now we can determine the speed of mass M and m just after they stick together in a perfectly inelastic collision. To do this, we will use conservation of linear momentum. The total momentum before the collision can be written as the sum of the individual momenta of mass M and mass m:
M_i = M*v_M + m*v_m
where M_i is the initial total momentum, v_M is the speed of mass M just before the collision (which is 0), and v_m is the speed of mass m just before the collision. The total momentum after the collision can be written as:
M_f = (M + m)*v_f
where M_f is the final total momentum and v_f is the speed of mass M and m just after they stick together in a perfectly inelastic collision. Since momentum is conserved, we have:
M_i = M_f
Substituting the expressions for M_i and M_f, we get:
M * 0 + m * v_m = (M + m) * v_f
Now we can solve this equation for v_f:
v_f = m * v_m / (M + m)
v_f = 0.200 * 2.426 / (0.300 + 0.200)
v_f ≈ 0.9704 m/s
So the speed of mass M and m just after they stick together in a perfectly inelastic collision is approximately 0.9704 m/s.
A light, ideal spring with a spring constant k = 100 N/m and uncompressed length L = 0.30 m is mounted to the fixed end of a frictionless plane inclined at an angle θ = 30.0° as shown above. Then a mass M 0.300 kg is affixed to the end of the spring and allowed to slowly slide down the plane until it is resting in an equilibrium position.
A second mass (m = 0.200 kg) is placed at a distance of 0.30 m above the equilibrium position (xo) of the mass-spring system. The second mass (m) is released so that it slides down the ramp and collides with the first mass (M).
a. Calculate the speed of mass, m just before it strikes mass, M.
b. Determine the speed of mass, M and m just after they stick together in a perfectly inelastic collision.
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