Physics Linear Momentum

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MALUBAG, ABEGAIL M.

PROBLEMS:

1. Four billiard balls, each of mass .5 kg, all are travelling in the same direction on a billiard table, with speeds 2 m/s, 4 m/s, 8 m/s and 10 m/s. What is the linear momentum of this system?

Given:
m = .5 kg
v1 = 2 m/s
v2 = 4 m/s
v3 = 8 m/s
v4 = 10 m/s
p = ?

Solution:
The linear momentum of a system is simply the sum of the linear momentum of the constituent parts. Thus we only need to find the momentum of each ball:

P = m 1 v 1 + m 2 v 2 + m 3 v 3 + m 4 v 4  = 1 + 2 + 4 + 5
= 12
Thus the total momentum of the system is 12 kg-m/s.

2. A 2500 kg bus from Laguna moves at 25 m/s. What is the momentum of the bus?

Given:
m = 2500 kg
v = 25 m/s
p = ?
Solution:
p = mv
= (2500kg)(25m/s)
= 62 500 kg-m/s

Thus the momentum of the bus is 62 500 kg-m/s.

3. A 20 kg boy is riding 6 kg bicycle with a velocity of 4m/s to the North. What is the total momentum of the boy and the bicycle together?

Given:
m = 26 kg
v = 4 m/s
p = ?
Solution:
p = mv
= (26kg)(4m/s)
= 104 kg-m/s

Thus the total momentum of the boy and the bicycle together is 104 kg-m/s.

MALUBAG, ABEGAIL M.

4. Which has a greater momentum: a bowling ball having a mass of 5 kg moving at 0.4 m/s or a baseball having a mass of 0.2 kg moving at 10 m/s?

Given:
m1 = 5 kg
m2 = 0.2 kg
v1 = 0.4 m/s
v2 = 10 m/s
p = ?

Solution:
The magnitude of the momentum of the bowling ball is:
p1 = m1v1
= (5kg)(0.4m/s)
= 2 kg-m/s

The magnitude of the momentum: a bowling ball is:
p2 = m2v2
= (0.2kg)(10m/s)
= 2 kg-m/s
The two balls have the same momentum of 2 kg-m/s.

5. A system is made up of three bodies with their respective velocities: body A of mass 1.5 kg And moving east at 2.0 m/s; body B of mass 2.0 kg moving west at 3.0 m/s; and body C of mass 5.2 kg moving west at 2.5 m/s. What is the momentum of the system?

Given:
m1 = 1.5 kg
m2 = 2 kg
m3 = 5.2 kg
v1 = 2 m/s
v2 = 3 m/s
v3 = 2.5 m/s
p = ?

Solution:
p1 = m1v1
= (1.5kg)(2m/s)
= 3 kg-m/s, east

p2 = m2v2
= (2kg)(3m/s)
= 6 kg-m/s, west

p3 = m3v3
= (5.2kg)(2.5m/s)
= 13 kg-m/s, west
The resultant momentum is 16 kg-m/s, west.

MALUBAG, ABEGAIL M.

6. A 300 kg bus speeding eastward along a super highway slows down from 6m/s to 4 m/s. What is the corresponding change in momentum?

Given:
m = 300 kg
Vi = 6 m/s
Vf = 4 m/s
Δp = ?
Solution:
Δp = m(Vf)-m(Vi)
= (300kg)(4kg) - (300kg) (6m/s)
= -600 kg-m/s

The corresponding change in momentum is 16 kg-m/s.

7. How fast must Superman run to have the same momentum as a 2500 kg train moving at 4.5 m/s in the same direction? Assume that the mass of Superman is 120 kg.

Given:
m1 = 2500 kg
m2 = 120 kg
v = 4.5 m/s
p = ?

Solution:
p = mv
= (2500kg)(4.5m/s)
= 11 250 kg-m/s

v = p/m
= 11 250 kg-m/s/120 kg
= 93.75 m/s
The momentum is 11 250 kg-m/s and so Superman must have a speed of 93.45 m/s.

8. What is the resultant momentum of a system of two particles with these masses and velocities; 3.75 kg moving north at 5.60 m/s and 4.20 kg moving northwest at 2.30 m/s?

Given:
m1 = 3.75 kg
m2 = 4.20 kg
Vi = 5.60 m/s
Vf = 2.30 m/s
p = ?

Solution:
p1 = m1v1
= (3.75kg)(5.60m/s)
= 21 kg-m/s, east

p2 = m2v2
= (4.20kg)(2.30m/s)
= 6 kg-m/s, west

MALUBAG, ABEGAIL M.

9. A 1.0 kg ball travelling at 4.0 m/s strikes a wall and bounces straight back at 2.0 m/s.

Given:
m = 1.0 kg
Vi = 4.0 m/s
Vf = ?
Vf = -2.0 m/s (opposite direction)
Δp = ?

Solution:
Δp = m(Vf-Vi)
= (1.0 kg)(-2.0 m/s - 4.0 m/s)
= -6 kg-m/s

J = Ft = mΔv =Δp
J = +6.0 kg-m/s
J = +6.0 kg-m/s

The momentum is -6 kg-m/s, west. The impulse applied to the ball is - 6.0 kg-m/s and...
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