**Kinematics**

**Problem 1:**

A ball that is thrown upward near the surface of the earth with a velocity of 50 m/s will come to rest about 5 second later. If the ball were thrown up with the same initial velocity on Planet X, after 5 seconds it would still be moving upwards at nearly 31 m/s. The magnitude of the gravitational field near the surface of Planet X is what fraction of the gravitational field near the surface of the earth?

Solution:

Concepts: One-dimensional motion with constant acceleration | |

Reasoning: We are given the velocity at t = 0 and t = 5 s and are asked to find the acceleration. | |

Details of the calculation: v(t) = v _{0} + at = v_{0} - g't. 31 m/s = 5 0 m/s - g'(5 s).g' = 3.8 m/s ^{2}. g'/g = 0.388. |

**Problem 2:**

A projectile launched from the ground explodes into three fragments of equal mass at the top point of the trajectory. One of the fragments lands t seconds after the explosion; two other fragments land simultaneously 2t seconds after the explosion. How high above the ground does the projectile explode?

Solution:

Concepts: Momentum conservation, kinematics | |

Reasoning: Let h be the height above the ground at which the projectile explodes and v _{jy}
(j = 1, 2, 3) the y-component of the velocity at the moment of the explosion
for the three fragments. Label with 1 the fragment which lands after t
seconds.Since the landing time depends only on the y-component of the velocity at the moment of the explosion, we have v _{3y} = v_{2y} .We then need only the equations for fragments 1 and 2, stating that at the landing time the y-coordinate will be zero. h + v _{1y}t – ½gt^{2} = 0, h + 2v_{2y}t – 2gt^{2}
= 0. | |

Details of the calculation: Since the masses of the three fragments are equal, and the y-component of the projectile momentum at the moment of the explosion is equal to 0, momentum conservation requires v _{1y }+ v_{2y} + v_{3y} = v_{1y }+ 2v_{2y}
= 0, v_{2y} = -½v_{1y}.We have 3 equations for 3 unknowns, v _{1y}, v_{2y}, and h.h + v _{1y}t – ½ gt^{2} = 0, h - v_{1y}t – 2gt^{2}
= 0.v _{1y}t – ½ gt^{2} = -v_{1y}t – 2gt^{2}, v_{1y}
= -¾gt.h = ¾gt ^{2} + ½gt^{2} = (5/4)gt^{2}. |

**Problem 3:**

Two points, A and B, are located on the ground a certain
distance d apart. Two rocks are launched simultaneously from points A and
B, with equal speeds but at different angles. Each rock lands at the
launch point of the other. Knowing that one of the rocks is launched at an
angle θ_{0} > 45^{o}, find the minimum distance between the
rocks during the flight in terms of d and θ_{0}?

Solution:

Concepts: Kinematics, projectile motion | |

Reasoning: We have to find the position of both rocks as a function of time. The range of a projectile launched with speed v _{0} at
an angle θ_{0} with respect to the horizontal x-axis is d = v_{0}^{2}sin(2θ_{0})/g.For two projectile launched with the same speed v _{0} to have the
same range they have to be launched at angles θ_{±} = π/4 ± θ. | |

Details of the calculation: Assume rock 1 is thrown from point A with speed v _{0 }and launch angle θ _{+} = θ_{0} = π/4 + θ towards
point B. Let the origin of the coordinate system be at point A, let the x-axis point towards B and let the y-axis point upward. For rock 1 we have x _{1}(t) = v_{0}cos(π/4
+ θ)t, y_{1}(t) = v_{0}sin(π/4
+ θ)t – ½gt^{2}.Assume rock 2 is thrown from point with speed v _{0 }and launch angle θ_{-}
= π/4 - θ towards point A. For rock 2 we have x _{2}(t)
= d - v_{0}cos(π/4 - θ)t,
y_{2}(t) = v_{0}sin(π/4 - θ)t
– ½gt^{2}.Let the distance between the rocks be L(t). L ^{2}(t) = (x_{2}(t) - x_{1}(t))^{2}
+ (y_{2}(t) - y_{1}(t))^{2}= v _{0}^{2}t^{2}(d/(v_{0}t)
- cos(π/4 - θ) - cos(π/4
+ θ))^{2} + v_{0}^{2}t^{2}(sin(π/4
- θ) - sin(π/4 + θ))^{2}= v _{0}^{2}t^{2}(d/(v_{0}t)
- cos(π/4)cos(θ) - sin(π/4)sin(θ)
– cos(π/4)cos(θ) + sin(π/4)sin(θ))^{2}
+ v _{0}^{2}t^{2}(sin(π/4)cos(θ)
- cos(π/4)sin(θ) – sin(π/4)cos(θ) - cos(π/4)sin(θ))^{2}
= 4v _{0}^{2}t^{2}(d/(2v_{0}t) - cos(π/4)cos(θ))^{2}
+ 4v_{0}^{2}t^{2}cos^{2}(π/4)sin^{2}(θ)= d ^{2} + 4v_{0}^{2}t^{2}cos^{2}(π/4)(cos^{2}(θ)
+ sin^{2}(θ)) - 4v_{0}^{2}t^{2}dcos(π/4)cos(θ)/(v_{0}t)= d ^{2} + 2v_{0}^{2}t^{2} -
4v_{0}tdcos(π/4)cos(θ).To find the time at which L ^{2}(t) has a minimum, set dL ^{2}(t)/dt = 4v_{0}^{2}t - 4v_{0}dcos(π/4)cos(θ)
= 0.t _{min} = dcos(π/4)cos(θ)/v_{0}.L ^{2}(t_{min}) = d^{2} + d^{2}cos^{2}(θ)
- 2d^{2}cos^{2}(θ) = d^{2} -
d^{2}cos^{2}(θ).L _{min}
= dsin(θ) = dsin(θ_{0} - π/4) is the minimum
distance between the rocks. |

**Problem 4:**

A catapult set on the ground can launch a rock a maximum horizontal distance
L. What would be the maximum horizontal launch distance if the catapult is set
on a platform moving forward with constant speed equal to the launch speed of the
rock?

Neglect the air resistance and assume that the rock is launched from the ground
level in both cases.

Solution:

Concepts: Kinematic equations, projectile motion | |

Reasoning: We have motion with constant acceleration. | |

Details of the calculation: Let θ _{0} be the launch angle with respect to the horizontal. (v_{x}
= v cosθ_{0}, v_{y} = v sinθ_{0})The range of a projectile is R = (v _{0}^{2}sin2θ_{0})/g.For the stationary catapult: R _{max} = v_{0}^{2}/g
= L, v_{0} = (gL)^{1/2}. (θ_{0} = 45^{o})For the moving catapult: R = (v _{0}^{2}sin2θ_{0})/g
+ v_{0}t. t = 2v _{y}/g = 2v_{0}sinθ_{0}/g.R = (v _{0}^{2}sin2θ_{0})/g + 2v_{0}^{2}sinθ_{0}/g,dR/dθ _{0} = 0 --> cos2θ_{0} = -cosθ_{0. }θ_{0}
= 60^{0}.R = (v _{0}^{2}/g)[sin2θ_{0} + 2sinθ_{0}]
= L 3√3/2. |