find the expression for the velocity, the speed, and how far in meters could the pin slide before stopping.

Question

A bowling ball of mass m_b = 3.9 kg is rolled down the lane with a velocity of v₁ = 2.1 m/s. It strikes a single remaining pin m_p = 0.89 kg head on. After the collision the bowling ball has a velocity (in the same direction) of v₂ = 1.6 m/s.

Part (a)  Write an expression for the velocity of pin in the horizontal direction after the impact, v_p. 

Part (b)  What is the pins speed in m/s? 

Part (c)  Assuming the coefficient of friction is μ_k = 0.1, how far in meters could the pin slide before stopping if it doesn't topple? ?

Answer 1

Let

m_b:mass of ball and m_p: mass of pin

v₁: initial velocity of ball and v₂: final velocity of ball

v_i: initial velocity of pin and v_f: final velocity of pin

a) From the conservation of momentum: momentum before collision equals momentum after collision

Therefore,

m_bv₁ + m_pv_i = m_bv₂ + m_pv_f ; since initial velocity of the pin is zero, the (final) velocity of the pin is

v_f = (m_b/m_p)(v₁ - v₂)

b) plugging in the known masses and velocities, you get the velocity of pin to be

v_f = 2.19 m/s

c)The frictional force, f_k = µ_kN = µ_km_pg = m_pa [where "a" is acceleration and N is normal force]

==> a = -µ_kg = -.1(9.8m/s²) = - 0.98m/s²

With this acceleration value (ignore the negative sign since we care only about the magnitude), you can calculate how far the pin will slide before stopping using

d = v_f²/2a = 2.45m

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You can equally find the time, t =v_p/a = 2.23s, and then use d = .5at² = 2.45m