You throw a 0.15 kg ball straight upward from ground level with an initial speed of 15 m/s. (Yes, from ground level. This is to make the calculations easier. If it bothers you, imagine you are standing in a dugout.) Your friend throws a 0.09 kg lump of clay upward after it. The ball is at a height of 10 m, coming back down, when the clay, going 25 m/s upward at that instant, strikes the ball and sticks to it. Assume the interaction is instantaneous, and neglect air resistance. (The easiest way to do this problem is from energy and momentum laws, not kinematics!)
(a)What is the velocity of the ball/clay combination immediately after the collision?
For this one, I used m1vi1+m2vi2=(m1+m2)vf
vf=30.09 m/s
(b)What is the speed of the combination right before it strikes the earth?
I don't know what equation I could use without using kinematics?
(c)How fast, and in which direction, would a bird have to be flying for the momentum of the system
to appear to her to be zero right before the collision?
b. You know the mass, velocity, starting height of the clay/ball combo. compute time to hit earth
you can use a number of equations, but if you think, on the way down at 10m, it will have the same velocity but down...
vf^2=vi^2+2ad
Vf^2=30.09^2+9.8*2*10
c. I don't understand the q.
To solve part (b) of the problem, you can use the principle of conservation of mechanical energy. The initial kinetic energy of the ball/clay combination is the sum of the kinetic energies of the ball and the clay separately before the collision. The final kinetic energy of the combination is the sum of the kinetic energy right before it strikes the earth.
The initial kinetic energy of the ball/clay combination is given by:
KE_initial = (1/2) * m1 * v1^2 + (1/2) * m2 * v2^2
Where m1 and m2 are the masses of the ball and clay respectively, and v1 and v2 are their respective initial velocities.
The final kinetic energy of the combination is given by:
KE_final = (1/2) * (m1 + m2) * V^2
Where (m1 + m2) is the total mass of the ball/clay combination, and V is the velocity of the combination right before it strikes the earth.
Since there is no loss of mechanical energy in the absence of external forces like air resistance, we can equate the initial and final kinetic energies:
KE_initial = KE_final
Substituting the given values, we get:
(1/2) * m1 * v1^2 + (1/2) * m2 * v2^2 = (1/2) * (m1 + m2) * V^2
Plugging in the values:
(1/2) * 0.15 kg * (15 m/s)^2 + (1/2) * 0.09 kg * (25 m/s)^2 = (1/2) * (0.15 kg + 0.09 kg) * V^2
Simplifying and solving for V, we would find the speed of the combination right before it strikes the earth.
To solve part (c), the momentum of the system must be zero right before the collision. Consider the momentum of the system in the vertical direction. The upward momentum of the ball/clay combination is equal to the downward momentum of the bird. If the bird is flying with a velocity that cancels out the momentum of the ball/clay combination, the total momentum of the system would appear to be zero to the bird right before the collision. The direction of the bird's velocity would depend on the direction of the initial velocity of the ball/clay combination.