Download Examples and problems to the system of particles

Examples and problems to the system of particles
Examples
1. Consider a system shown in the figure. The masses of the blocks are respectively
m1 = 3 kg, m2 = 8 kg. The coefficient of kinetic friction is µ = 0,25. The mass of the
pulley can be neglected.
a) After the blocks are released from rest, find the acceleration of the system and the
tension in the cord.
b) Find the speed of the system after moving s = 1,5 m.
m2
m1
Solution:
a) The figure shows the forces exerted on the blocks: (K is the tension in the rope, in the
English textbooks it’s notation is usually T.)
F
n
m2
F
K
2
f
K1
a
m1
G2
G1
Now, for a massless string, if the mass of the pulley is neglected, the tension is the same
everywhere along the string and is the same at each end of it: K 1 = K 2 = K
Applying Newton’s second law to each of the masses:
m1a = m1 g − K
m2 a = K − F f
Eliminating K and writing the magnitude of the frictional force, as F f = µFn = µm2 g ,
m1 g − µm2 g
= 0,91m / s 2
m1 + m2
Substituting the result into one of the previous equations, we obtain: K = 27,27 N
b) To calculate the final speed we can use the work-energy law:
(m1g − µm2 g )s = 1 (m1 + m2 )v 2
2
(m g − µm2 g ) s = 1,65 m / s
Hence
v= 2 1
m1 + m2
We obtain
a=
2. In the next figure the masses of the pulley can be negligible and there is no friction. Find
the accelerations of the blocks.
m1
m2
Solution:
The figure shows the forces and the accelerations:
1
Fn
m1
a1
K1
K1
G1
K1
K2
a2
K2
m2
G2
Applying Newton’s second law to each of the masses:
m1 a1 = K 1
m2 a 2 = m2 g − K 2
The relationship between the tensions is: K2 = 2K1, because of the negligible masses of the
pulleys.
The relationship between the acceleration can be found from the magnitude of the
displacements of the blocks in the same time interval:
s2 = (1/2)a2 t2, s2 = s1/2
s1 = (1/2)a1 t2,
a
Hence
a2 = 1
2
Therefore from these equations
m2 g
m2 g
2m 2 g
a1 =
=
; a2 =
m
4m1 + m2
4m1 + m2
2m1 + 2
2
3. The figure shows a ballistic pendulum with mass m2 = 10 kg, for measuring the speed of a
bullet with mass m1 = 0,05 kg. The initial speed of the bullet is v1 = 200 m/s, the length of
the pendulum l =2 m. The bullet is fired at the block and becomes embedded in it.
a) Find the common speed immediately after the impact.
b) Find the dissipated energy during the collision.
c) Find the maximum value of the angular velocity.
Solution:
a) The conservation of momentum for the inelastic collision:
m1v1 + m2v2 = (m1 + m2 )v'
Hence
v’ = 0,995 m/s
b) The magnitude of the dissipated energy comes from the change of the kinetic energy:
1
1
∆Wkin = (m1 + m2 )v'2 − m1v12 = −995 J
2
2
c) After the collision, for the block-bullet system, according to the conservation of the
mechanical energy
1
(m1 + m2 )v'2 = (m1 + m2 )ghmax
2
Hence
hmax = 0,0495 m
Using the next figure we can calculate the angular displacement.
Φmax
l
hmax
2
hmx = l − l cos ϕ max
cos ϕ max = 0,975,
o
Φmax = 12,77 .
4. A bowl of mass m1 =4 kg, with coming with speed v1 = 2 m/s, collides with a bowl of
mass m2 = 6 kg, which is initially at rest. After the collision the first bowl has a velocity
v1’ = 1 m/s making an angle of α1 =53o with the initial direction. Find the velocity of the
second bowl after the collision. Calculate the dissipated energy.
Solution:
To use the conservation of momentum in the collision, make a drawing for the momentums:
I1 '
α1
I1
α2
I1
I2 '
Now we can apply the cosine law for the collision, as the conservation of momentum:
I 2 '2 = I1 '2 + I12 − 2 I1 ' I1 cos α1
I 2 ' = m2v2 '
Hence
v2 ' = 1,07 m/s
The direction of the velocity comes from the direction of linear momentum. Thus from the ycomponents of the momentum after the collision:
I1 ' sin α1 = I 2 ' sin α 2
m v ' sin α1
sin α 2 = 1 1
= 0,5
m2v2 '
α 2 = 30o
The dissipated energy in the collision comes from the change of the kinetic energy:
∆Wkin =
1
1
1
m1v1 '2 + m2v2 '2 − m1v12 = −2,565 J
2
2
2
Problems
1. In the arrangement the strings and pulleys are massless and there is no friction. The
masses are respectively m1 = 3 kg, m2 = 2 kg, m3 = 2 kg. Find the acceleration of the system
and the tensions in the strings. (a = 1,43 m/s2, K1 = 25,7 N, K2 = 22,86 N)
m2
m3
m1
2. A block of mass m2 = 2 kg on a horizontal table is connected to a mass m1 = 1 kg by a
string which passes over a small frictionless pulley. The coefficient friction between
the table and the mass is µ = 0,2.
a) Find the acceleration of the system and the tension in the string. (a = 2 m/s2; K = 8 N)
b) Find the speed of the system after s = 1,5 m. (v = 2,45 m/s)
3. The coefficients of kinetic friction between the M = 1,5 kg block and the slope with
α = 30o is µk = 0,12. If m = 2 kg, determine the magnitude and direction of the
acceleration and the tension in the string. The string and pulley is massless. (a = 3,13
m/s2, up the slope, K = 13,75 N)
3
M
m
α
4. A boy pulls two sleds which are connected by a string with F = 120 N in α= 40 ° to
the horizontal. The first sled has a mass m1 = 16 kg, and the second sled has a mass
m2 = 17 kg. The coefficient of friction is µ = 0,02. Find the acceleration of the sleds.
What is the tension in the string between the two sleds? (a = 2,63 m/s2; K = 48,15 N)
F
m2
m1
α
5. Two crates (m1 = 300 kg, m2 = 100 kg) joined by a rope, are being pulled along a
factory floor by a constant force exerted on the m1 at an angle α= 25° to the
horizontal. The µ1 = 0,11 (coefficient of kinetic friction between the heavier crate and
the floor) and µ2 = 0,18 (coefficient of kinetic friction between the lighter crate and the
floor). What should the magnitude of the applied force be to move the crates at
constant speed? What applied force would be required to give the system an
acceleration of a = 0,1 m/s2? What would be the tension in the rope connecting the two
crates in the latter case? (F1 = 535,7 N; F2 = 577,7 N; K = 190 N)
6. A bullet of mass m = 13 g has a horizontal velocity and strikes and becomes
embedded in a block of mass M = 12 kg, which can swing on a string. The length of
the string is l = 1,5 m. The maximum angle with the vertical axis for the combined
mass of the block and bullet is α = 11o.What was the speed of the bullet before
impact? (v1= 686 m/s)
7. A ball of mass m1 = 0,5 kg moving at v1 = 2 m/s and a ball of mass m2 = 0,3 kg
moving at the same speed in opposite direction, collides head-on. The collision is
elastic. Find the magnitude and direction of their velocities after the collision. (v1’ = -1
m/s, v2’ = 3 m/s)
8. A 2000 kg car and a 1000 kg car traveling at 40 m/s in opposite directions collide
head-on and lock together. What are their speed and direction right after the collision?
(v’ = 13,3 m/s)
9. The particle of mass m1 =0,1 kg has the speed before the collision v1 and the ball of
m2 = 0,2 is at rest. After the collision the ball of m1scattered with the angle of 45o and
velocity of v1’ = 1,41 m/s. The angle of the velocity of m2 after the collision is 30o to
the v1. Find the speed of the m2 after the collision. Calculate the velocity of m1 before
the collision. Give the loss in mechanical energy. (v2’ = 1 m/s; v1 = 2,73 m/s; ∆Wk = 0,173 J)
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