Mass M_1 = 2.25 kg is at the end of a rope that is 2.00 m in length. The initial
ID: 3278096 • Letter: M
Question
Mass M_1 = 2.25 kg is at the end of a rope that is 2.00 m in length. The initial angle with respect to the vertical is 60.0 degree and M_1 is initially at rest. Mass M_1 is released and strikes M_2 = 4.50 kg exactly horizontally. The collision is elastic. After collision, mass M_2 is moving on a frictionless surface, but runs into a rough patch 2.00 m in length, with coefficients of friction = 0.175 for both kinetic and static. At the end of the rough patch, the object is again on a frictionless surface and is stopped by a spring with spring constant equal to k = 50.0 N/m. Calculate the following: a) Velocities of Masses M_1 and M_2 after collision b) The amount the spring is compressed while stopping mass M_2. c) What maximum height after collision does mass M_1 obtain? Simulate on IP and determine the % errors in the quantities found above.Explanation / Answer
initial height of the M1 = 60 degree wrt to vertical
mass, M1 = 2.25 kg
length of rope = l = 2m
initial potential energy of M1 = M1*g*l(1 - cos(60)) = 2.25*g*2(1 - cos(60)) = 22.0725 J
just before striking at the bottom most point of the trajectory, the KE of the M1 = PE at heighest point ( from conservation of energy)
so, let velocity of M1 just before striking be u
then 0.5M1u^2 = 22.0725 J
u = 4.429 m/s
now before collision velocity of M2 = 0 m/s
mass ,M2 = 4.5 kg
After collision, let velocity og M1 be v1 and thet of M@ be v2
from conservation of momentum
M1u = M1v1 + M2v2 => 2.25*4.429 = 2.25*v1 + 4.5*v2
2.25*v1 + 4.5*v2 = 9.966
also, from the fact that this was an elastic collsion
so coefficient of restitution = 1
(v2 - v1)/(u1 - u2) = 1
(v2 - v1) = 1*4.429
v2 = 4.429 + v1
a. so, v1= [9.966 - 4.5*4.429]/(2.25 + 4.5) = -1.476 m/s
v2 = 2.95277 m/s
b. KE of M2 after collision = 0.5M2*v2^2 = 0.5*4.5*v2^2 = 19.617 J
friction force on the rough patch, f = mu*M2*g = 0.175*4.5*g = 7.72 N
energy lost top friction = f*d = 2f = 15.45 J
energy left in M2 after rough patch = 4.16625 J ( this energy is transferred to the spring)
maximum compression of spring be x
then 0.5kx^2 = 4.16625
0.5*50*x^2 = 4.16625
x = 0.40822 m
c. M1 velocity after collision, v1 = -1.476 m/s
KE of M1 at this point = 0.5*M1*v1^2 = 2.45 J = PE at heighest point = M1*gh
h = 0.111 m