1.
Picture the Problem: A baseball pitcher gives the ball momentum when he throws it.
Strategy: Multiply the mass and the speed to calculate the magnitude of the momentum.
Solution: Calculate the momentum:
p  mv   0.142 kg  45.1 m/s  6.40 kg  m/s
Insight: This momentum will change slightly due to the effects of gravity, as we will shortly learn, but not by
much because the pitch will cross the plate less than half a second after leaving the pitcher’s hand!
2.
Picture the Problem: A dog has momentum as it races to meet its owner.
Strategy: Use the definition of momentum to find the dog’s speed.
Solution: Solve the definition of momentum for v:
p  mv
v
p 37 kg  m/s

 3.1 m/s
m
12 kg
Insight: The dog’s speed is equivalent to 6.9 mi/hr. Either the dog isn’t “racing” after all, or it’s a really small
dog for whom 6.9 mi/hr is a fast run!
3.
Picture the Problem: The speeds and masses of four objects are given.
Strategy: Multiply the mass and the speed of each object to find its momentum, and then use the results to rank
the momenta of the four objects.
pA  mA vA  10 kg 10 m/s   100 kg  m/s
Solution: 1. Calculate the momenta:
pB  mB vB  15 kg  4 m/s   60 kg  m/s
pC  mC vC   5 kg  20 m/s   100 kg  m/s
pD  mD vD   60 kg  3 m/s   180 kg  m/s
2. Compare the momenta to arrive at the ranking: pB < pA = pC < pD .
Insight: The momenta of objects A and C could be different even though their magnitudes are the same
because momentum is a vector, and the directions of these momentum vectors weren’t specified.
6.
Picture the Problem: One car drives due north and an identical car drives due east with the same speed.
Strategy: The total momentum of the two vehicles is the vector sum of their individual momenta. Find the
magnitude of the momentum of each car and then add the two momentum vectors.
Solution: 1. Find the
magnitude of the momentum:
2. Add the components of the
two momenta:
p  mcar vcar  1200 kg15 m/s  18,000 kg  m/s
p total  p car 1  pcar 2
ptotal, x  pcar 1, x  pcar 2, x  0  18, 000 kg  m/s 
ptotal, y  pcar 1, y  pcar 2, y  18, 000 kg  m/s   0
3. Find the total momentum:
ptotal 

2
2
pcar
1, x  pcar 1, y
18, 000 kg  m/s  2  18, 000 kg  m/s  2
 25, 000 kg  m/s
p total  2.5  104 kg  m/s at 45 north of east
Insight: The answer is written in scientific notation in order to emphasize that it has only two significant digits.
The 45° direction can either be inferred or calculated:


  tan 1 ptotal, y ptotal, x  tan 1 18,000 kg  m/s 18,000 kg  m/s  45.
7.
If the mass of an object doubles, its momentum also doubles if the speed remains the same, because p  mv.
8.
Because the momentum vector is given by p  m v, and because mass is not a vector, then it is clear that the
momentum vector must point in the same direction as the velocity vector. The two sides of a vector equation
such as p  m v must be equal in both magnitude and direction.
9.
If the speed of an object doubles, the magnitude of its momentum will double because p  m  2v   2 mv.
However, its kinetic energy will quadruple because KE  12 m  2v   4  12 mv 2 . Therefore, the momentum
doubles but the kinetic energy quadruples.
2
10. The total momentum of two moving objects can be zero because momentum is a vector, and two equal and
opposite momenta will cancel each other. However, kinetic energy is a scalar quantity, and two moving objects
will each have nonzero, positive kinetic energy that cannot cancel. Therefore, it does not follow that if a system
has zero momentum it must have zero kinetic energy as well.
11. Picture the Problem: An apple moves in a straight line at a known speed.
Strategy: Multiply the mass and the speed to calculate the magnitude of the momentum.
Solution: Calculate the momentum:
p  mv   0.16 kg  2.7 m/s  0.43 kg  m/s
Insight: The momentum of the apple will not change unless a net force acts upon it.
12. Picture the Problem: A school bus and a baseball each have nonzero momentum.
Strategy: Set the momentum of the school bus equal to the momentum of the baseball and solve for the
baseball’s speed.
Solution: Solve for the speed of the baseball:
pbus  pball
mbus vbus  mball vball
mbus vbus
18, 200 kg 12.5 m/s 
 vball 
mball
 0.142 kg 
vball  1, 600, 000 m/s  1.60  106 m/s
Insight: The baseball’s speed would be 4,670 times faster than the speed of sound!
13. Picture the Problem: A dog has momentum as it runs.
Strategy: Use the definition of momentum to find the dog’s mass.
p 17 kg  m/s

 8.9 kg
v
1.9 m/s
Insight: A dog of this mass would weigh 20 lb on Earth’s surface, so this is a medium-sized dog.
Solution: Solve the definition of momentum for m:
p  mv
m
14. Picture the Problem: Two different carts approach each other on a frictionless track at different speeds.
Strategy: Add the momenta of the two carts, taking the momentum of cart 1 to be positive.
Solution: Add the two momenta:
p total  m1 v1  m2 v 2
  0.35 kg 1.2 m/s    0.61 kg   0.85 m/s 
p total  0.42    0.52 kg  m/s   0.10 kg  m/s
Insight: If cart 2 were traveling at 0.69 m/s, its momentum would exactly cancel the momentum of cart 1 and
the total momentum of the system would be zero.