Question

A 3000-kg truck moving with a velocity of 1- m/s hits a 1000-kg parked car. The impact causes the 1000-kg car to be set in motion at 15 m/s. Assuming that momentum is conserved during the collision, determine the velocity of the truck immediately after the collision.

Answer 1

Answer:

given that :

The mass of truck(m1) = 3000kg

velocity of truck(u1) = 1m/s

The mass of car(m2) = 1000kg

velocity of car(u2) = 0m/s

After collision,

velocity of truck(v1) = v

velocity of car(v2) = 15m/s

we know that,

m1u1 + m2u2 = m1v1 + m2v2

(3000*1)kg.m/s + (1000*0)kg.m/s =

(3000*v)kg + (1000*15)kg.m/s

3000kg.m/s = 3000kg*v +15000kg.m/s

-12000kg.m/s = 3000kg*v

v = (12000/3000)m/s

v = -4m/s

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Answer 2

The velocity of the truck after the collision is $-4m/s$.

Step by step explanation

Given data:

Mass of the truck$=3000kg$

Velocity of the truck$=1m/s$

mass of the car$=1000kg$

velocity of the car after the impact $=15m/s$

To find:

We must determine the velocity of the truck soon after collision.

Formula to be used:

$m_{1} u_{1} +m_{2} u_{2} =m_{1} v_{1} +m_{2} v_{2}$

where:$m_{1} =$ $3000kg$ (mass of the truck)

$m_{2}=$$1000kg$ (mass of the car)

$u_{1} =$$1m/s$ (initial velocity of the truck)

$u_{2} =$ $0$ (initial velocity of the car)

$v_{1} =$final velocity of the truck

$v_{2}=$$15m/s$ (final velocity of the car)

Solution:

Using the formula:$m_{1} u_{1} +m_{2} u_{2} =m_{1} v_{1} +m_{2} v_{2}$

$3000$×$1+1000$×$0=3000$×$v_{1} +1000$×$15$

$3000=3000v_{1}+15000$

$v_{1}$$=\frac{3000-15000}{3000}$

$v_{1} =-4m/s$

Therefore the velocity of the truck after the collision is $-4m/s$.

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