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Semenov [28]
3 years ago
14

) a 1.0 kilogram laboratory cart moving with a velocity of 0.50 meter per second due east collides with and sticks to a similar

cart initially at rest. after the collision, the two carts move off together with a velocity of 0.25 meter per second due east. the total momentum of this frictionless system is

Physics
2 answers:
ZanzabumX [31]3 years ago
4 0
Momentum would be the same before and after the collision
 Before the collision:
 Momentum of the single cart: 1 * 0.50 = 0.50
 After the collision
 velocity = 0.25m / s
 1 * 0.25 + 1 * 0.25 =
 0.25 * (1 + 1) =
 0.25 * 2 =
 0.50
 Now new momentum will be 0.5
 answer
 the same before and after the collision
nirvana33 [79]3 years ago
3 0

The momentum of the system of two cars after the collision will be \boxed{0.50\text{ kg.m/s}}.

Explanation:

Given:

The mass of each of the laboratory car is 1\text{ kg}.

The velocity of the laboratory cart is 0.50\text{ m/s}.

Concept:

The momentum of a body is defined as the amount of motion contained in the body at the particular instant. The momentum of the body is defined as the product of the mass and velocity of the body.

Write the expression for the momentum of the laboratory cart.

p_{1}=m\times{v}                                               ...... (1)

Here, p_1 is the momentum of the cart, m is the amss of the cart and v is the velocity of the cart.

The initial momentum of the laboratory cart which is in motion.

\begin{aligned}p_{i}&=(1\text{ kg})\times{(0.5\text{ m/s})}\\&=0.5\text{ kg.m/s}\end{aligned}

As the cart collides and the surface is frictionless, the momentum of the system will remain conserved after the collision.

The two carts will stick to one another after the collision. So, the mass of the system will be 2\text{ kg} after the collision and the speed of the carts after collision is 0.25\text{ m/s}.

Substitute the value of mass and velocity of the system in the equation of momentum.

\begin{aligned}p'&=(2\text{ kg})\times(0.25\text{ m/s})\\&=0.50\text{ m/s}\end{aligned}

Thus, the momentum of the system of two cars after the collision will be \boxed{0.50\text{ kg.m/s}}.

Learn More:

1. The net force acting on the helicopter at the position brainly.com/question/6125929

2. Shortest time taken by car to accelerate from 0 to 80 mph brainly.com/question/7174363

Answer Details:

Grade: High School

Subject: Physics

Chapter: Momentum

Keywords:

laboratory cart, 0.50 m/s, east, collides, momentum, conserved, total momentum, similar cart, sticks, velocity.

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A box is being moved with a velocity (v) by a force P (parallel to v) along a level horizontal floor. The normal force is (Fn),
labwork [276]

Answer:

Force (P) : Positive

Normal Force (Fn) : Zero

Weight (mg) : Zero

Kinetic Frictional Force (fk) : Negative

Explanation:

The work done by a force on an object is given by the following formula:

W = F.d

W = F d Cosθ

where,

W = Work Done

f = Force Applied

d = displacement

θ = Angle between force and displacement

<u>FOR FORCE (P)</u>:

Since, force P is parallel to the motion of the box. Therefore, θ = 0°

Hence,

W = P d Cos 0°

W = P d(1)

W = Pd

<u>Therefore, work done by force (P) is Positive.</u>

<u></u>

<u>FOR NORMAL FORCE (Fn) AND WEIGHT (W)</u>:

Since, normal force and weight are perpendicular to the motion of the box. Therefore, θ = 90°

Hence,

W = Fn d Cos 90°= mg d Cos 90°

W = Fn d(0) = mg d (0)

W = 0

<u>Therefore, work done by Normal Force (Fn) and Weight (mg) is Zero.</u>

<u></u>

<u>FOR KINETIC FRICTIONAL FORCE (fk)</u>:

Since, kinetic frictional force acts in the opposite direction of motion of the box. Therefore, θ = 180°

Hence,

W = fk d Cos 180°

W = fk d(-1)

W = -fk d

<u>Therefore, work done by Kinetic Frictional Force (fk) is Negative.</u>

<u></u>

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