Question

A trolley with mass m1 = 0.8 kg and speed v1 = 4.0 m / s collide with another stationary trolley with mass m2 = 4.0 kg. Immediately after the collision, carriage 1 bounces back at the speed u1 = 0.5 m / s.
a) Calculate the speed of trolley 2 immediately after collision 0 / CP / AB
(b) What proportion of trolley 1 movement energy is converted to other forms of energy during the collision 0 / CP / 0

Answer 1

Given,

$$m_1=0.8\: kg,m_2=4.0\: kg,v_1=4.0m/s,v_2=0,u_1=0.5m/s$$

a).Let speed of Trolley 2 after collision is $u_2$ .

Now, by momentum conservation we get,

$$m_1v_1+m_2v_2=m_1u_1+m_2u_2$$

Thus,$0.8\times4.0+0=0.8\times 0.5+4.0u_2\implies u_2=0.7m/s$

b).During the collision speed of both objects are same thus, again applying momentum conservation we get,

$$m_1v_1=(m_1+m_2)v\implies v\approx0.67m/s$$

Now. initial energy was ,

$$K_i=\frac{1}{2}m_1v_1^2=6.4J$$

Energy during the collision on the system is

$$K_f=\frac{1}{2}(m_1+m_2)v^2=1.08J$$

Thus, energy which converted to other forms is

$$K_i-K_f=5.32J$$