Question

4) If an object of 40 kg is moving east at 20 m/s and collides with a 20 kg object moving along the same vector at 10 m/s, what will be the velocity of the smaller object if the larger object slows to 10 m/s?

Answer 1

Answer:

$v_2_2=30m/s$

Explanation:

In all collisions the total linear momentum of the system is conserved. Therefore:

$p_i=p_f\\\\Where:\\\\p_i=m_1_1 v_1_1+m_2_1 v_2_1\\\\p_f=m_1_2 v_1_2+m_2_2 v_2_2$

So, $p_i$ represents the linear momentum before the collision and $p_f$ represents the linear momentum after the collision. Now, let:

$m_1_1=Mass\hspace{3} of\hspace{3} the\hspace{3} bigger\hspace{3} o bject\hspace{3}before\hspace{3}the\hspace{3}collision\\v_1_1=Velocity \hspace{3} of\hspace{3} the\hspace{3} bigger\hspace{3} o bject\hspace{3}before\hspace{3}the\hspace{3}collision\\m_2_1=Mass\hspace{3} of\hspace{3} the\hspace{3} smaller\hspace{3} o bject\hspace{3}before\hspace{3}the\hspace{3}collision\\v_2_1=Velocity \hspace{3} of\hspace{3} the\hspace{3} smaller\hspace{3} o bject\hspace{3}before\hspace{3}the\hspace{3}collision$

$m_1_2=Mass\hspace{3} of\hspace{3} the\hspace{3} bigger\hspace{3} o bject\hspace{3}after\hspace{3}the\hspace{3}collision\\v_1_2=Velocity \hspace{3} of\hspace{3} the\hspace{3} bigger\hspace{3} o bject\hspace{3}after\hspace{3}the\hspace{3}collision\\m_2_2=Mass\hspace{3} of\hspace{3} the\hspace{3} smaller\hspace{3} o bject\hspace{3}after\hspace{3}the\hspace{3}collision$

$v_2_2=Velocity \hspace{3} of\hspace{3} the\hspace{3} smaller\hspace{3} o bject\hspace{3}after\hspace{3}the\hspace{3}collision$

According to the data provided by the problem:

$m_1_1=40kg\\v_1_1=20m/s\\m_2_1=20kg\\v_2_2=10m/s\\m_1_2=40kg\\v_1_2=10m/s\\m_2_2=20kg\\v_2_2= \text{?}$

Replacing the data into the linear momentum equation and solving for $v_2_2$:

$m_1_1 v_1_1+m_2_1v_2_1=m_1_2 v_1_2+m_2_2 v_2_2\\\\(40)(20)+(20)(10)=(40)(10)+20(v_2_2)\\\\800+200=400+20(v_2_2)\\\\20(v_2_2)=600\\\\ (v_2_2)=\frac{600}{20} =30m/s$

Thus, the velocity of the smaller object  after the collision is 30m/s

Answer 2

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.

$m_{1}v_{1} + m_{2}v_{2}=m_{1}v_{3}+m_{2}v_{4}$

where $m_{1}, m_{2} , v_{1}and v_{2}$ are the masses  and velocities of the objects 1 and 2 respectively before collision.

$40\times 20 + 20\times 10 = 40\times 10 + 20\times v_{4}$

$1000= 400 + 20v_{4}$

$v_{4}=30m/s$

Hence, the object will move at 30 m/s