Question:
In pairs figure-skating competition, a 65 kg man and his 45 kg female partner stand facing each other on skates on the ice. If they push apart and the woman has a velocity of 1.5 m/s eastward, what is the velocity of her partner? (neglect friction)
Solution:
Given values:
mass of man = 65 kg velocity of man = ?
mass of woman = 45 kg velocity of woman = 1.5 m/s
M1V1 = M2V2 This equation is used when 2 values of a type and 1 value of another type are
65kg(Vm) = (45kg)(1.5m/s) given. Conservation of momentum is used when 2 objects collide, the
Vm = (45kg)(1.5m/s)/(65kg) momentum before equals the momentum after.
Vm = 1.0 m/s Her partner takes off with a velocity of 1.0 m/s in the westward direction.
Given values:
mass of man = 65 kg velocity of man = ?
mass of woman = 45 kg velocity of woman = 1.5 m/s
M1V1 = M2V2 This equation is used when 2 values of a type and 1 value of another type are
65kg(Vm) = (45kg)(1.5m/s) given. Conservation of momentum is used when 2 objects collide, the
Vm = (45kg)(1.5m/s)/(65kg) momentum before equals the momentum after.
Vm = 1.0 m/s Her partner takes off with a velocity of 1.0 m/s in the westward direction.
Question:
To get off a frozen, frictionless lake, a 65 kg person takes of a .150 kg shoe and throws it horizontally, directly away from the shore with a speed of 2 m/s. if the person is 5m from the shore, how long does he take to reach it?
Solution:
Given values:
mass of man = 65 kg velocity of man (Vm) = ? Δx = 5m
mass of shoe = .15 kg velocity of shoe = 2 m/s
M1V1 = M2V2 x = VmT
65kg(Vm) = .15kg(2m/s) T = x/Vm
Vm = .0046153 m/s = 5/.0046153
t = 1083s
The conservation of momentum formula is used to find the velocity the man travels at by plugging in the values given to us in the problem. There is no acceleration in the horizontal direction because the ice is frictionless, so we use the formula Δ x=Vot. After plugging in the values for Δx and the velocity of the man, we get 1083 s is the amount of time the man takes to get back to shore.
Given values:
mass of man = 65 kg velocity of man (Vm) = ? Δx = 5m
mass of shoe = .15 kg velocity of shoe = 2 m/s
M1V1 = M2V2 x = VmT
65kg(Vm) = .15kg(2m/s) T = x/Vm
Vm = .0046153 m/s = 5/.0046153
t = 1083s
The conservation of momentum formula is used to find the velocity the man travels at by plugging in the values given to us in the problem. There is no acceleration in the horizontal direction because the ice is frictionless, so we use the formula Δ x=Vot. After plugging in the values for Δx and the velocity of the man, we get 1083 s is the amount of time the man takes to get back to shore.
Question:
A 1.0 kg ball is thrown horizontally with a velocity of 13 m/s and the contact time is .020 s, what force is exerted on the ball by the wall?
Solution:
Given values:
m = 1.0 kg final velocity (Vf) = 13 m/s Force = ?
Δt = 0.020 s initial velocity (Vi) = 15 m/s
Δp = mΔv Δp = mΔv: This equation means the change in momentum is mass times the change in velocity.
F = Δp/Δt The equation for average force is used. We substitute the change in momentum formula into the
= mΔv/Δt impulse formula.
= (1)(15-(-13))/(0.020) We plug in the values and get the force exerted on the ball by the wall is equal to 1400 N.
= 1400 N
Given values:
m = 1.0 kg final velocity (Vf) = 13 m/s Force = ?
Δt = 0.020 s initial velocity (Vi) = 15 m/s
Δp = mΔv Δp = mΔv: This equation means the change in momentum is mass times the change in velocity.
F = Δp/Δt The equation for average force is used. We substitute the change in momentum formula into the
= mΔv/Δt impulse formula.
= (1)(15-(-13))/(0.020) We plug in the values and get the force exerted on the ball by the wall is equal to 1400 N.
= 1400 N