Problem 1 An object with m1 = 5kg is attached to a spring of negligible mass. This mass/spring combination is then slid horizont
ally on a frictionless surface with a velocity of 5 m/s towards a stationary object with m2 = 6kg. Upon impact, the spring compresses, then we examine two cases. First, find the velocities of the two objects assuming the spring completely relaxes again after the interaction. Second, assume that m2, after they separate, slides up a frictionless incline. (a)What is the relative speed of the masses when the spring is maximally compressed?
1 answer:
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Answer:
Explanation:
Given that,
x component of a vector = -12 m
The y component of a vector = -15 m
We need to find the direction of a vector. The direction of a vector is given by :
Put all the values,
So, the direction of vector is to x component.
Answer:
The required angle is (90-25)° = 65°
Explanation:
The given motion is an example of projectile motion.
Let 'v' be the initial velocity and '∅' be the angle of projection.
Let 't' be the time taken for complete motion.
Let 'g' be the acceleration due to gravity
Taking components of velocity in horizontal(x) and vertical(y) direction.
= v cos(∅)
= v sin(∅)
We know that for a projectile motion,
t =
Since there is no force acting on the golf ball in horizonal direction.
Total distance(d) covered in horizontal direction is -
d = ×t = vcos(∅)×
=
.
If the golf ball has to travel the same distance 'd' for same initital velocity v = 23m/s , then the above equation should have 2 solutions of initial angle 'α' and 'β' such that -
α +β = 90° as-
d = =
=
=
.
∴ For the initial angles 'α' or 'β' , total horizontal distance 'd' travelled remains the same.
∴ If α = 25° , then
β = 90-25 = 65°
∴ The required angle is 65°.