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A car of mass 400 kg and travelling at 72 km/ hr crashes into a truck of mass 4000 kg and travelling at 9 km/ hr in same direction The car bounces back at speed 18 km/ hr The speed of truck after the impact is-. a. 9 km/ hr. b 18 km/ hr. c. 27 km/ hr?
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A car of mass 400 kg and travelling at 72 km/ hr crashes into a truck ...
Let mass of car m1= 400 kg,
Initial velocity of car u1=70km/hr =19.44 m/s
Let the mass of truck m2=4000 kg
Initial velocity of truck u2 = 9km/hr =2m/s
Final velocity of car v1=18km/hr = 5m/s
 velocity of truck v2 =?
According to law of conservation of momentum. Total momentum before collision is equal to total momentum after collision.

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A car of mass 400 kg and travelling at 72 km/ hr crashes into a truck ...
The problem involves a collision between a car and a truck. We are given the masses and initial velocities of both vehicles, as well as the final velocity of the car after the collision. We need to determine the final velocity of the truck after the impact.

Given:
- Mass of car (m1) = 400 kg
- Mass of truck (m2) = 4000 kg
- Initial velocity of car (v1i) = 72 km/hr
- Initial velocity of truck (v2i) = 9 km/hr
- Final velocity of car (v1f) = -18 km/hr (since it bounces back)

To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

1. Calculate the initial momentum before the collision:
- Momentum of car before the collision (p1i) = m1 * v1i
- Momentum of truck before the collision (p2i) = m2 * v2i

2. Calculate the final momentum after the collision:
- Momentum of car after the collision (p1f) = m1 * v1f
- Momentum of truck after the collision (p2f) = m2 * v2f (to be determined)

3. Apply the conservation of momentum:
- Total momentum before the collision (p_total_initial) = p1i + p2i
- Total momentum after the collision (p_total_final) = p1f + p2f

According to the conservation of momentum:
p_total_initial = p_total_final

4. Substitute the values and solve for v2f:
m1 * v1i + m2 * v2i = m1 * v1f + m2 * v2f

Now let's convert the velocities from km/hr to m/s:
- v1i = 72 km/hr * (1000 m/3600 s) = 20 m/s
- v2i = 9 km/hr * (1000 m/3600 s) = 2.5 m/s
- v1f = -18 km/hr * (1000 m/3600 s) = -5 m/s

Substituting the values:
400 kg * 20 m/s + 4000 kg * 2.5 m/s = 400 kg * -5 m/s + 4000 kg * v2f

5. Solve for v2f:
(400 kg * -5 m/s) - (400 kg * 20 m/s) = 4000 kg * v2f - (4000 kg * 2.5 m/s)
-2000 kg m/s - 8000 kg m/s = 4000 kg * v2f - 10000 kg m/s
-10000 kg m/s - 4000 kg * v2f = -10000 kg m/s

Simplifying the equation:
-4000 kg * v2f = 0
v2f = 0 m/s

Therefore, the final velocity of the truck after the impact is 0 m/s. This means that the truck comes to a complete stop after the collision with the car.
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A car of mass 400 kg and travelling at 72 km/ hr crashes into a truck of mass 4000 kg and travelling at 9 km/ hr in same direction The car bounces back at speed 18 km/ hr The speed of truck after the impact is-. a. 9 km/ hr. b 18 km/ hr. c. 27 km/ hr? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A car of mass 400 kg and travelling at 72 km/ hr crashes into a truck of mass 4000 kg and travelling at 9 km/ hr in same direction The car bounces back at speed 18 km/ hr The speed of truck after the impact is-. a. 9 km/ hr. b 18 km/ hr. c. 27 km/ hr? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A car of mass 400 kg and travelling at 72 km/ hr crashes into a truck of mass 4000 kg and travelling at 9 km/ hr in same direction The car bounces back at speed 18 km/ hr The speed of truck after the impact is-. a. 9 km/ hr. b 18 km/ hr. c. 27 km/ hr?.
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