Question

The truck A of mass 3500 kg, travelling at 50 m/s, collides with another truck B of mass 3000 kg
travelling at 45 m/s in the same direction. After collision the velocity of Truck A becomes 40 m/s.
Calculate the velocity of truck B after collision.
PLEASE ANSWER FAST​

Answer 1

Given : -

• Mass of truck A = 3500kg
• Initial velocity of truck A = 50m/s
• Mass of truck B = 3000kg
• Initial velocity of truck B = 45m/s
• Both are moving in the same direction.
• After collision, velocity of truck A becomes 40m/s.

To Find : -

• Velocity of truck B after collision.

Solution : -

In the case of collision, no external force acts on the whole system so linear momentum of system is conserved.

We know that momentum is measured as the product of mass and velocity.

Initial momentum of the system will be equal to the final momentum of the system.

$\longrightarrow \sf P_{initial}= P_{final} \\\\\longrightarrow m_1u_1+m_2u_2 = m_1v_1+m_2v_2 \\\\\longrightarrow ( 3500 \times 50)+( 3000 \times 45)= (3500 \times 40)+3 \times v_2 \\\\\longrightarrow (3.5 \times 50)+( 3 \times 45)=(3.5 \times 40) +3 \times v_2 \\\\\longrightarrow 175+135= 140+3v_2 \\\\\longrightarrow 3v_2 = 310 - 140 \\\\\longrightarrow 3v_2 = 170 \\\\\longrightarrow\bf v_2 = 56.67 \ m/s$

Velocity of truck B = 56.67 m/s²

Answer 2

Given:

• Mass$\sf{(M_1)}$ of Truck A= 3500kg

• Mass$\sf{(M_2)}$ of Truck B=3000kg

• Initial Velocity$\sf{(U_1)}$ of Truck A=50m/s

• Initial Velocity$\sf{(U_2)}$ of Truck B=45m/s

• Final Velocity$\sf{(V_1)}$ of Truck A=40m/s.

• Both Trucks are moving in same direction.

To Find:

Final Velocity of Truck B

Solution:

According to the law of conservation of momentum,

"If two objects collide, then the total momentum before and after the collision will be the same if there is no external force acting on the colliding objects."

Mathematically,

$\sf{P_{initial}=}{P_{final}}$

Therefore,

$\sf{\implies}{m_1 u_1 +m_2 u_2 = m_1 v_1 +m_2 v_2}$

$\sf{\implies}{(3500 \times 50)+(3000 \times 45)=(3500 \times 40)+(3000 \times v_2)}$

$\sf{\implies}{(1000) \times (175)+(1000)\times (135)=(1000)\times (140)+(1000)\times (3v_2)}$

$\sf{\implies}{(1000)\times(175+135)=(1000)\times(140+3v_2)}$

$\sf{\implies}{310=140 \times 3v_2}$

$\sf{\implies}{3v_2=310-140}$

$\sf{\implies}{3v_2=170}$

$\sf{\implies}{v_2=}{\dfrac{170}{3}}$

$\sf{\implies}{v_2=56.6ms^{-1}}$

Thus, the Velocity of Truck B = 56.6m/s