A child's game consists of a block that
attaches to a table with a suction cup, a
spring connected to that block, a ball, and a
launching ramp. By compressing the
spring, the child can launch the ball up the
ramp. The spring has a spring constant k,
the ball has a mass m, and the ramp raises a
height h. The spring is compressed a
distance S in order to launch the ball. When
the ball leaves the launching ramp its
velocity makes an angle θ with respect to the horizontal.
(a) Calculate the velocity of the ball when it just leaves the launching ramp (both magnitude and
direction. Be sure to specify your coordinate system.)
(b) The spring constant = 1000.0 N/m, the spring's compression is 4.00 cm, the ball's mass is
55.0 grams, the height of the ramp is 15.0 cm, and the top of the table is 1.20 m above the floor.
With what total speed will the ball hit the floor? (Use g = 10.0 m/s2)
4 answers
SpringcompressedEnergy + initialPE
=KE final
The initial PE is mg*1.35
solve for the total KE when it hits the floor, and from that, its speed.
I will be happy to check your work.
1/2ks^2+mgh+1/2mv^2 = 0
solving for v.. i get
v = sqrt(-(2gh-ks^2/m)
thats what is confusing me. so im not shure if i did it correctly.
=KE final
1/2 k s^2+mgh=1/2 m v^2 This v is the velocity at the floor when it hits. h is the intial height of the ball (table). h is 1.20, not 1.35 as I indicated.
v=sqrt(2gh+ks^2/m)
The above is just sayingthe intial energy has to equal the final energy, and initial energy is spring energy + initial potential energy (mg*1.2)