Question

In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The collision is cushioned by a spring (k = 1200 N/m).

a) Determine the velocity of each cart after the collision.
b) Determine the maximum compression of the spring.

Answer 1

The answer is given below

Step-by-step explanation:

u is the initial velocity, v is the final velocity. Given that:

`m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m`

a)

The final velocity of cart 1 after collision is given as:

`v_1=((m_1-m_2)/(m_1+m_2))u_1+(2m_2)/(m_1+m_2)u_2\\ Substituting:\\v_1=(0.6-0.8)/(0.6+0.8) (-5)+(2*0.8)/(0.6+0.8)(2)= 5/7+16/7=3\ m/s`

The final velocity of cart 2 after collision is given as:

`v_2=((m_2-m_1)/(m_1+m_2))u_2+(2m_1)/(m_1+m_2)u_1\\ Substituting:\\v_1=(0.8-0.6)/(0.6+0.8) (2)+(2*0.6)/(0.6+0.8)(-5)= 2/7-30/7=-4\ m/s`

b) Using the law of conservation of energy:

`(1)/(2)m_1u_1+ (1)/(2)m_2u_2=(1)/(2)m_1v_1+(1)/(2)m_2v_2+(1)/(2)kx^2\\x=\sqrt{(m_1u_1+m_2u_2-m_1v_1-m_2v_2)/(k)}\\ Substituting\ gives:\\x=\sqrt{(0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2))/(1200)}=√(0)=0\ cm`