Question

After being struck by a bowling ball, a 1.8 kg bowling pin sliding to the right at 5.0 m/s collides head-on with another 1.8 kg bowling pin initially at rest. Find the final velocity of the second pin in the following situations: a) The first pin moves to the right after the collision at 0.8 m/s. Answer in units of m/s. 010 (part 2 of 2) 10.0 points b) The first pin stops moving when it hits the second pin. Answer in units of m/s

Answer 1

Answer:

a) v₂ = 4.2 m/s

b) v₂ = 5 m/s

Explanation:

a)

We will use the law of conservation of momentum here:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0.8 m/s

v₂ = speed of second after before collsion = ?

Therefore,

$(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0.8\ m/s)+(1.8\ kg)(v_2)\\v_2 = 5\ m/s - 0.8\ m/s$

v₂ = 4.2 m/s

b)

We will use the law of conservation of momentum here:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0 m/s

v₂ = speed of second after before collsion = ?

Therefore,

$(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0\ m/s)+(1.8\ kg)(v_2)$

v₂ = 5 m/s