Question

After you pick up a spare, your bowling ball rolls without slipping back toward the ball rack with a linear speed of 2.85 m/s. To reach the rack, the ball rolls up a ramp that rises through a vertical distance of 0.53 m. What is the linear speed of the ball when it reaches the top of the ramp?

Answer 1

Answer:

V' = 0.84 m/s

Explanation:

given,

Linear speed of the ball, v = 2.85 m/s

rise of the ball, h = 0.53 m

Linear speed of the ball, v' = ?

rotation kinetic energy of the ball

$KE_r = \dfrac{1}{2}I\omega^2$

I of the moment of inertia of the sphere

$I = \dfrac{2}{5}MR^2$

 v = R ω

using conservation of energy

$KE_r = \dfrac{1}{2}( \dfrac{2}{5}MR^2)(\dfrac{V}{R})2$

$KE_r = \dfrac{1}{5}MV^2$

Applying conservation of energy

Initial Linear KE + Initial roational KE = Final Linear KE + Final roational KE + Potential energy

$\dfrac{1}{2}MV^2 + \dfrac{1}{5}MV^2 = \dfrac{1}{2}MV'^2 + \dfrac{1}{5}MV'^2 + M g h$

$0.7 V^2 = 0.7 V'^2 + gh$

$0.7\times 2.85^2 = 0.7\times V'^2 +9.8\times 0.53$

V'² = 0.7025

V' = 0.84 m/s

the linear speed of the ball at the top of ramp is equal to 0.84 m/s

Answer 2

The linear speed of the ball when it reaches the top of the ramp is 0.84 m/s.

The given parameters;

• initial linear speed of the ball, u = 2.85 m/s
• vertical distance traveled by the ball, h = 0.53 m

Apply the principle of conservation of energy to determine the linear speed of the ball when it reaches the top of the ramp.

$K.E_i + P.E_i = K.E_f + P.E_f \\\\(K.E + K.E_r)_i + P.E_i = (K.E + K.E_r)_f \ + P.E_f\\\\(\frac{1}{2} mv^2 + \frac{1}{2} I \frac{V^2}{R^2} )_i + 0 = (\frac{1}{2} mv^2 + \frac{1}{2} I \frac{V^2}{R^2} )_f + mgh$

where;

• I is moment of inertia of the spherical ball

$(\frac{1}{2} mv^2 + \frac{1}{2} \times \frac{2}{5} mR^2\times \frac{v^2}{R^2} )_i = (\frac{1}{2} mv^2 + \frac{1}{2} \times \frac{2}{5} MR^2\times \frac{v^2}{R^2} )_f + mgh\\\\(\frac{1}{2} mv^2 + \frac{1}{5} mv^2)_i = (\frac{1}{2} mv^2 + \frac{1}{5} mv^2)_f + mgh\\\\0.7 mv_i^2 = 0.7mv_f^2+ mgh\\\\0.7 v_i^2 = 0.7v_f^2+ gh\\\\0.7v_f^2 = 0.7v_i^2 -gh\\\\v_f= \sqrt{\frac{0.7v_i^2 -gh}{0.7} } \\\\v_f = \sqrt{\frac{0.7(2.85)^2 -9.8(0.53)}{0.7} }\\\\v_f = 0.84 \ m/s$

Thus, the linear speed of the ball when it reaches the top of the ramp is 0.84 m/s.

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