A golfer hits a golf ball with a club head velocity of 94 mph. Mass of golf club head (m): 190 g Mass of golf ball (ms): 46 g Co
1 answer:
Answer:
Velocity of golf ball≅ 12.87mph
Explanation:
Using the theory of conservative momentum to both the club head and the golf ball we have;
Qi1 + Qi2= Qf1 + Qf2
Qi1: initial momentum for the club head
Qf1: final momentum for the club head
Qi2: initial momentum for the golf ball
Qf2: final momentum for the golf ball
momentum (Q) = Mass x velocity
which means the sum of the momentums of both club head and golf ball has to be the same before and after they have collided.
using the Coeficient of restitution e= 0.83 allows us to know what kind of collision we are dealing with, which is a partially elastic collision since
0> e=0.83 >1.
Qi1= 190 x 94= 17,860 mph.g
Qf1= 190 x Vf1
Qi2= 46 x 0= 0 mph.g
Qf2= 46x Vf2
Using the value of e to determine Vf2 as the final velocity of the golf ball:
e=
0.83=
Vf1= (0.83 x 94)+ Vf2
Vf1= 78.02 + Vf2
Qi1 + Qi2= Qf1 + Qf2
17,860 + 0 = 190xVf1 + 46xVf2
17,860= 190x (78.02+Vf2) + 46xVf2
17,860= 14,823+ 190xVf2+46xVf2
Vf2≅ 12.87mph
You might be interested in
Answer:
a) F_net = 6.48 10⁻¹⁸ ( ), b) x = 0.15 m
Explanation:
a) In this problem we use that the electric force is a vector, that charges of different signs attract and charges of the same sign repel.
The electric force is given by Coulomb's law
F =
Since when we have the two negative charges they repel each other and when we fear one negative and the other positive attract each other, the forces point towards the same side, which is why they must be added.
F_net= ∑ F = F₁ + F₂
let's locate a reference system in the load that is on the left side, the distances are
left side - electron r₁ = x
right side -electron r₂ = d-x
let's call the charge of the electron (q) and the fixed charge that has equal magnitude Q
we substitute
F_net = k q Q ( )
F _net = kqQ ( )
let's substitute the values
F_net = 9 10⁹ 1.6 10⁻¹⁹ 4.50 10⁻⁹ ( )
F_net = 6.48 10⁻¹⁸ ( )
now we can substitute the value of x from 0.05 m to 0.25 m, the easiest way to do this is in a spreadsheet, in the table the values of the distance (x) and the net force are given
x (m) F (N)
0.05 27.0 10-16
0.10 8.10 10-16
0.15 5.76 10-16
0.20 8.10 10-16
0.25 27.0 10-16
b) in the adjoint we can see a graph of the force against the distance, it can be seen that it has the shape of a parabola with a minimum close to x = 0.15 m
