m a = 10 g = 0.01 k g m_a= 10g=0.01kg m a = 10 g = 0.01 k g
m b = 20 g = 0.02 k g m_b= 20g=0.02kg m b = 20 g = 0.02 k g
v ⃗ a = − v ⃗ b \vec v_a= -\vec v_b v a = − v b
according to the law of conservation of momentum: \text{according to the law of conservation of momentum:} according to the law of conservation of momentum:
m a v ⃗ a + m b v ⃗ b = ( m a + m b ) v ⃗ m_a\vec v_a+m_b\vec v_b=(m_a+m_b)\vec v m a v a + m b v b = ( m a + m b ) v
− m a v ⃗ b + m b v ⃗ b = ( m a + m b ) v ⃗ -m_a\vec v_b+m_b\vec v_b=(m_a+m_b)\vec v − m a v b + m b v b = ( m a + m b ) v
v ⃗ = m b − m a m a + m b v ⃗ b \vec v=\frac{m_b-m_a}{m_a+m_b}\vec v_b v = m a + m b m b − m a v b
v ⃗ = v ⃗ b 3 \vec v=\frac{\vec v_b}{3} v = 3 v b
before collision \text{before collision} before collision
E k = m a v a 2 2 + m b v b 2 2 E_k= \frac{m_av_a^2}{2}+\frac{m_bv_b^2}{2} E k = 2 m a v a 2 + 2 m b v b 2
E k = m a v b 2 2 + m b v b 2 2 E_k= \frac{m_av_b^2}{2}+\frac{m_bv_b^2}{2} E k = 2 m a v b 2 + 2 m b v b 2
E k = v b 2 ( m a + m b ) 2 = 0.015 v b 2 E_k= \frac{v_b^2(m_a+m_b)}{2}=0.015v_b^2 E k = 2 v b 2 ( m a + m b ) = 0.015 v b 2
after collision: \text{after collision:} after collision:
E k ′ = v 2 ( m a + m b ) 2 = v b 2 ( m a + m b ) 18 = 0.01 v b 2 6 E'_k= \frac{v^2(m_a+m_b)}{2}=\frac{v_b^2(m_a+m_b)}{18}=\frac{0.01v_b^2}{6} E k ′ = 2 v 2 ( m a + m b ) = 18 v b 2 ( m a + m b ) = 6 0.01 v b 2
p = E k − E k ′ E k = 1 − E k ′ E k p = \frac{E_k-E'_k}{E_k}=1-\frac{E'_k}{E_k} p = E k E k − E k ′ = 1 − E k E k ′
p = 1 − 0.01 v b 2 6 ∗ 0.015 ∗ v b 2 ≈ 0.89 = 89 % p = 1- \frac{0.01v_b^2}{6*0.015*v_b^2}\approx0.89=89 \% p = 1 − 6 ∗ 0.015 ∗ v b 2 0.01 v b 2 ≈ 0.89 = 89%
Answer: 89% of kinetic energy is lost \text{Answer: 89\% of kinetic energy is lost} Answer: 89% of kinetic energy is lost
Comments