Answer:
The smaller object will move at 30 m/s
Explanation:
From the principle of conservation of linear momentum,
Momentum of objects before collision = momentum of objects after collision.
[tex]m_{1}v_{1} + m_{2}v_{2}=m_{1}v_{3}+m_{2}v_{4}[/tex]
where [tex]m_{1}, m_{2} , v_{1}and v_{2}[/tex] are the masses and velocities of the objects 1 and 2 respectively before collision.
[tex]40\times 20 + 20\times 10 = 40\times 10 + 20\times v_{4}[/tex]
[tex]1000= 400 + 20v_{4}[/tex]
[tex]v_{4}=30m/s[/tex]
Hence, the object will move at 30 m/s
Answer:
[tex]v_2_2=30m/s[/tex]
Explanation:
In all collisions the total linear momentum of the system is conserved. Therefore:
[tex]p_i=p_f\\\\Where:\\\\p_i=m_1_1 v_1_1+m_2_1 v_2_1\\\\p_f=m_1_2 v_1_2+m_2_2 v_2_2[/tex]
So, [tex]p_i[/tex] represents the linear momentum before the collision and [tex]p_f[/tex] represents the linear momentum after the collision. Now, let:
[tex]m_1_1=Mass\hspace{3} of\hspace{3} the\hspace{3} bigger\hspace{3} o bject\hspace{3}before\hspace{3}the\hspace{3}collision\\v_1_1=Velocity \hspace{3} of\hspace{3} the\hspace{3} bigger\hspace{3} o bject\hspace{3}before\hspace{3}the\hspace{3}collision\\m_2_1=Mass\hspace{3} of\hspace{3} the\hspace{3} smaller\hspace{3} o bject\hspace{3}before\hspace{3}the\hspace{3}collision\\v_2_1=Velocity \hspace{3} of\hspace{3} the\hspace{3} smaller\hspace{3} o bject\hspace{3}before\hspace{3}the\hspace{3}collision[/tex]
[tex]m_1_2=Mass\hspace{3} of\hspace{3} the\hspace{3} bigger\hspace{3} o bject\hspace{3}after\hspace{3}the\hspace{3}collision\\v_1_2=Velocity \hspace{3} of\hspace{3} the\hspace{3} bigger\hspace{3} o bject\hspace{3}after\hspace{3}the\hspace{3}collision\\m_2_2=Mass\hspace{3} of\hspace{3} the\hspace{3} smaller\hspace{3} o bject\hspace{3}after\hspace{3}the\hspace{3}collision\\[/tex]
[tex]v_2_2=Velocity \hspace{3} of\hspace{3} the\hspace{3} smaller\hspace{3} o bject\hspace{3}after\hspace{3}the\hspace{3}collision[/tex]
According to the data provided by the problem:
[tex]m_1_1=40kg\\v_1_1=20m/s\\m_2_1=20kg\\v_2_2=10m/s\\m_1_2=40kg\\v_1_2=10m/s\\m_2_2=20kg\\v_2_2=$\text{?}[/tex]
Replacing the data into the linear momentum equation and solving for [tex]v_2_2[/tex]:
[tex]m_1_1 v_1_1+m_2_1v_2_1=m_1_2 v_1_2+m_2_2 v_2_2\\\\(40)(20)+(20)(10)=(40)(10)+20(v_2_2)\\\\800+200=400+20(v_2_2)\\\\20(v_2_2)=600\\\\ (v_2_2)=\frac{600}{20} =30m/s[/tex]
Thus, the velocity of the smaller object after the collision is 30m/s
