Answer:
8.75
Explanation:
First, find the force of friction.
Kinetic energy = work done by friction
½ mv² = Fd
½ (3.9 kg) (2.9 m/s)² = F (1.4 m)
F = 11.7 N
Next, find the distance at the new velocity.
Kinetic energy = work done by friction
½ mv² = Fd
½ (3.9 kg) (2.5 × 2.9 m/s)² = (11.7 N) d
d = 8.75 m
Answer:
a. 8.96 m/s b. 1.81 m
Explanation:
Here is the complete question.
a) A long jumper leaves the ground at 45° above the horizontal and lands 8.2 m away.
What is her "takeoff" speed v
0
?
b) Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0 m away horizontally and 2.5 m, vertically below.
If she long jumps from the edge of the left bank at 45° with the speed calculated in part a), how long, or short, of the opposite bank will she land?
a. Since she lands 8.2 m away and leaves at an angle of 45 above the horizontal, this is a case of projectile motion. We calculate the takeoff speed v₀ from R = v₀²sin2θ/g. where R = range = 8.2 m.
So, v₀ = √gR/sin2θ = √9.8 × 8.2/sin(2×45) = √80.36/sin90 = √80.36 = 8.96 m/s.
b. We use R = v₀²sin2θ/g to calculate how long or short of the opposite bank she will land. With v₀ = 8.96 m/s and θ = 45
R = 8.96²sin(2 × 45)/9.8 = 80.2816/9.8 = 8.192 m.
So she land 8.192 m away from her bank. The distance away from the opposite bank she lands is 10 - 8.192 m = 1.808 m ≅ 1.81 m
Answer:
a) 
b) 
Explanation:
Given:
mass of the lighter block, 
velocity of the lighter block, 
mass of the heavier block, 
velocity of the heavier block, 
a)
Using conservation of linear momentum:

where:
final velocity of the lighter block
final velocity of the heavier block

........................(1)
Since kinetic energy is conserved in elastic collision:



divide the above equation by eq. (1)
.............................(2)
now we substitute the value of v from eq. (2) in eq. (1)



(negative sign denotes that the direction is towards left)
b)
now we substitute the value of v' from eq. (2) in eq. (1)




Answer:
12.5 m/s
Explanation:
The motion of the hammer is a free fall motion, so a uniformly accelerated motion, therefore we can use the following suvat equation:

Where, taking downward as positive direction, we have:
s = 8 m is the displacement of the hammer
u = 0 is the initial velocity (it is dropped from rest)
v is the final velocity
is the acceleration of gravity
Solving the equation for v, we find the final velocity:

So, the final speed is 12.5 m/s.