A child\'s game consists of a block that attaches to a table with a suction cup,

Question

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 ? 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 1.10 m above the floor. With what total speed will the ball hit the floor? (Use g = 10.0 m/s2) Physics 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 ? 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 1.10 m above the floor. With what total speed will the ball hit the floor? (Use g = 10.0 m/s2) Physics

Explanation / Answer

a) E1 (energy at spring, only spring potential)=1/2(k)x^2

= (1/2)(1000)(0.04^2)

= 0.80 J

E2 (top of ramp, both KE and PE)=mgh + 1/2(mv^2)

= 0.055(10)(0.15)+(1/2)(0.055)v^2

Conservation of energy due to no friction

E1=E2

0.80=0.0825+0.0275v^2

0.7175/0.0275=v^2

v=5.12 m/s @ top of ramp

b) E3 introduced: energy of ball on the floor = mgh + 1/2(mv^2)

E3=E1=E2

0.80=0.055(10)(-1.1) + (1/2)(0.055)v^2

1.405 = 0.0275v^2

51.09 = v^2

v = 7.12 m/s

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