Using a spring-loaded launcher, you measure the velociy of a 25 mm diameter stee

Question

Using a spring-loaded launcher, you measure the velociy of a 25 mm diameter steel sphere to be 2 m/s. The sphere has a mass of 66 g. You want to hit the sphere with a small blob of clay (mass 2 g), which will hit the metal sphere at right angles, if you time it right. You have a slingshot which can produce launch speeds of (at least) 30 m/s for such a projectile, which gives an effective spring constant of 720 N/m.

(a) You pull the slingshot back (from its equilibrium position) by 3 cm. Find the launch speed of the clay ball.

(b) The clay ball sticks to the steel. Find the direction and magnitude of the velocity of the combined objects in the two dimensions shown on the diagram, which is looking from above the experiment. Ignore, for now, the fact that both balls are also falling towards the ground.

(c) The steel ball travels 50 cm before the collision, and started 1 m off the ground. Find the final landing point (of the steel plus clay) relative to the launch point of the steel ball.

Explanation / Answer

(A) Applying energy conservation,

k x^2 / 2 + 0 = 0 + m v^2 / 2

720 (0.03^2) = (0.002) v^2

v = 18 m/s

(B) Applying momentum conservation,

(66 x 2 i) + (2 x 18 j) = (66 + 2) vf

vf = 1.94i + 0.53 j m/s

magnitude = sqrt(1.94^2 + 0.52^2) = 2.01 m/s

direction angle = tan^-1(0.53/1.94) = 15.3 deg

(c) in vertical,

y = yf - yi = 0 - 1 = -1 m

v0y = 0.53 m/s

ay = - 9.81 m/s^2

Applying y = v0y t + ay t^2 /2

- 1 = 0.53t - 4.9 t^2

t = 0.51 sec

d = 1.94 x 0.51 = 1 m

r = 1.5i - 1j

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