Answer to Question #185926 in Classical Mechanics for Kaitlyn Hosking

Question

1. During an experiment, a vehicle, mass 400 g, is pushed so that it runs along an air track with a velocity of 2.0 ms^-1. It collides with a stationary vehicle, mass 600 g, which moves on with a velocity of 1.6 ms^-1

Find the velocity of the 400 g vehicle after the collision.

1. A 2kg trolley travelling at 3 m/s hits a stationary 1kg trolley.  After the collision the 2kg trolley continues to travel in the same direction at 1 m/s. 

The 1 kg trolley moves off separately. 

Calculate the velocity of the 1kg trolley after the collision.

1. A truck of mass 2000kg, moving at 20m/s, collides with a car of mass 1000kg, which is stationary. What is the velocity of the car after the collision if the truck now moves 

forward at 10m/s ?

1. A skater of mass 70 kg is moving at 4 m/s when he bumps into a skater of mass 50 kg who is not moving.  They move off together on the ice.  What is their velocity immediately after the collision?

Answer 1

Solution 1.

$m_1=400g$

$u_1=2.0 m/s$

$m_2=600g$

$u_2=0$

After collision, $v_2=1.6m/s$

Now, applying the conservation of linear momentum,

$m_1 u_1+m_2u_2=m_1v_1+m_2v_2$

$\Rightarrow 400\times 2.0 +600\times 0 = 400\times v_1+600\times 1.6$

$\Rightarrow v_1=\frac{800-960}{400}m/s$

$\Rightarrow v_1=\frac{-160}{400}m/s$

$\Rightarrow v_1=- 0.4 m/s$

Here, negative sign indicates that after the collision, object will move in opposite direction.

Solution 2.

$m_1=2kg, m_2= 1kg$

$u_1=3m/s$

$u_2=0$

After the collision,

$v_1= 1m/s$

$v_2=?$

Now, applying the conservation of linear momentum,

$m_1 u_1+m_2u_2= m_1 v_1+m_2v_2$

$\Rightarrow 2\times 3+1\times 0 =2\times 1 +1\times v_2$

$\Rightarrow v_2=\frac{6-2}{1}= 4 m/s$

Solution 3.

$m_1= 2000kg$

$u_1=20 m/s$

$m_2= 1000kg$

$u_2=0$

After collision, $v_1= 10 m/s$

$v_2=?$

Now, applying the conservation of linear momentum,

$m_1 u_1+m_2u_2=m_1v_1+m_2v_2$

Now, substituting the values,

$2000\times 20+ 1000\times 0 = 2000\times 10+1000\times v_2$

$\Rightarrow v_2= \frac{40000-20000}{1000} m/s$

$\Rightarrow v_2= \frac{20000}{1000} m/s$

$\Rightarrow v_2= 20 m/s$

Solution 4.

$m_1= 70 kg$

$m_2=50 kg$

$u_1= 4m/s$

$u_2=0$

After the collision, both the two blocks stick,

Now, applying the conservation of linear momentum,

$m_1 u_1+m_2 u_2= (m_1+m_2) v$

Now, substituting the values,

$\Rightarrow 70 \times 4+50\times 0=(50+70)v$

$\Rightarrow v=\frac{280}{120}m/s$

$\Rightarrow v = 2.33 m/s$