(b) If the straight-line distance from her home to the university is 10.3 km in a direction 25.0° south of east, what was her av
1 answer:
You might be interested in
Answer:
Explanation:
As we know that backpack is kicked on the rough floor with speed "v"
So here as per force equation in vertical direction we know that
so normal force on the block is given as
now the magnitude of kinetic friction on the block is given as
now when bag is sliding on the floor then net deceleration of the block due to friction is given as
now we know that bag hits the opposite wall at L distance away in time t
so we have
Answer:
YOUR answer is given below:
Explanation:
Answer:
Speed of ball A after collision is 3.7 m/s
Speed of ball B after collision is 2 m/s
Direction of ball A after collision is towards positive x axis
Total momentum after collision is m×4·21 kgm/s
Total kinetic energy after collision is m×8·85 J
Explanation:
<h3>If we consider two balls as a system as there is no external force initial momentum of the system must be equal to the final momentum of the system</h3>Let the mass of each ball be m kg
v be the velocity of ball A along positive x axis
v be the velocity of ball A along positive y axis
u be the velocity of ball B along positive y axis
Conservation of momentum along x axis
m×3·7 = m× v
∴ v = 3.7 m/s along positive x axis
Conservation of momentum along y axis
m×2 = m×u + m× v
2 = u + v → equation 1
∴ relative velocity of approach = relative velocity of separation
-2 = v - u → equation 2
By adding both equations 1 and 2 we get
v = 0
∴ u = 2 m/s along positive y axis
Kinetic energy before collision and after collision remains constant because it is an elastic collision
Kinetic energy = (m×2² + m×3·7²)÷2
= 8·85×m J
Total momentum = m×√(2² + 3·7²)
= m× 4·21 kgm/s