What is the velocity of Tanya's car after the collision?

Question

Bill forgets to put his car in park and it starts rolling west. When it is moving at a speed of 3.5 m/s, it collides with Tanya's car, which is moving at a speed of 2 m/s east. Bill's car bounces backward (to the east) at a speed of 2 m/s. Bill's car has a mass of 900 kg, and Tanya's car has a mass of 1100 kg. i solved it but i donot know if it is right?

a. What is the velocity of Tanya's car after the collision? (2 points)

950 = 900*(-2)+1100*u

u=2.5 m/s

b. Assuming all lost kinetic energy is converted to heat, how much energy is converted to heat during the collision? (2 points)

ke before collision= 1/2 (900)(3.5)² + 1/2 (1100)(2)²=5512.5+2200=7712.5J

 ke after collision=1/2(900)(2)² + 1/2 (1100)(2.5)² = 1800+3437.5=5237.5 J

= (7712.5-5237.5) J

= 2475 J

Answer 1

A picture is worth a thousand words!

First, set up your frame of reference. This is important in establishing direction (which is a critical part of the answer). Remember that momentum is a vector composed of velocity (a vector) so the velocity answer must have both magnitude and direction. I chose my frame of reference to be that east is positive and west is negative.

P₁ = P₂ (Conservation of Momentum)

P₁ = (m_A1)(v_A1) + (m_B1)(v_B1)

P₁ = (1100)(2) + (900)(-3.5) = 2200 - 3150 = -950 kgm/s

P₂ = (m_A2)(v_A2) + (m_B2)(v_B2)

P₂ = (1100)(v_A2) + (900)(2) = (1100)(v_A2) + 1800

Using P₁ = P₂ we have -950 = (1100)(v_A2) + 1800

Solving for v_A2:

-2750/1100 = v_A2:or

v_A2:= -2.5 meaning 2.5 m/s in the west direction

So, you were correct except the direction was needed.

For part b) you are correct BECAUSE Kinetic Energy is NOT a vector so we do not need to worry about the sign of the velocity.

Good job!

Answer 2

Hi Laveeza,

I think you have the right idea but the answer is not correct. The momentum before the collision is:

(mass of Bill's car)*(its velocity) = (900 kg)*(3.5 m/s) = 3150 kg*m/s

The momentum after the collision is exactly what you said it is: (900kg*m/s)*(-2)+1100kg*u

So, by conservation of momentum we get the following equation:

(900kg*m/s)*(-2)+1100kg*u = 3150kg*m/s

Solving for u, we get:

1100kg*m/s * u = 4950kg*m/s

u = 4.5 m/s

Hopefully that makes sense. Let me know if you still need help with part b.