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# Tag Archives: conservation of momentum

## Is linear momentum conserved when a ball is thrown vertically up in the sky?

ROUNAK BHATTACHARJEE asked the above question.

The fact is that linear momentum is conserved in all interactions. If you concentrate on the motion of a single body alone, that will not account for an interaction. When thrown above, according to the principle of conservation of momentum, what we can say is that “The total momentum of the ball plus the thrower just before throwing is equal to the total momentum just after throwing”

After that what happens is controlled by the force of gravity acting on it. The force acting on it can change the momentum. Here the interaction is between the ball and the earth. On account of the huge mass of earth, the motion of earth to compensate the motion of ball and to conserve the momentum cannot be detected.

However, the linear momentum is conserved in all events and interactions so far!

## Numericals from Class XI Physics Conservation of Momentum)

- A 41.0-kg boy, riding a 1.60-kg skateboard at a velocity of 5.70 m/s across a level sidewalk, jumps forward to leap over a wall. Just after leaving contact with the board, the boy’s velocity relative to the sidewalk is 6.00 m/s, 9.30° above the horizontal. Ignore any friction between the skateboard and the sidewalk. What is the skateboard’s velocity relative to the sidewalk at this instant?
- A 1.20-g bullet, traveling at a speed of 478 m/s, strikes the wooden block of a ballistic pendulum. The block has a mass of 191 g. (a) Find the speed of the bullet/block combination immediately after the collision. (b) How high does the combination rise above its initial position?
- After skiding down a snow-covered hill on an inner tube, Ashley is coasting across a level snowfield at a constant velocity of +3.0 m/s. Miranda runs after her at a velocity of +4.2 m/s and hops on the inner tube. How fast do the two of them slide across the snow together on the inner tube? Ashley’s mass is 71 kg, and Miranda’s is 58 kg. Ignore the mass of the inner tube and any friction between the inner tube and the snow.
- A car (mass = 1170 kg) is traveling at 28.8 m/s when it collides head-on with a sport utility vehicle (mass = 2580 kg) traveling in the opposite direction. In the collision, the two vehicles come to a halt. At what speed was the sport utility vehicle traveling?
- In an Atwood system mass 1= 2 kg and mass 2=7kg. The masses of the pulley and strings are negligible by comparison, the pulley turns without friction and the string does not stretch. The lighter object is released with a sharp push that sets into motion at v(initial)= 2.4 m/s downward.

How far will m1 descend below it’s initial level?

Find the velocity of m1 after 1.8 seconds

## A few Numerical Problems from Mechanics

ninzyy posted:

PLEASE ANSWER AT LEAST ONE OF THESE QUESTIONS WITH EXPLANATIONS IF POSSIBLE. ITS DUE TOMORROW!!! PLEASE HELP!

**1. A 12 kg block sliding at 6.5 m/s along a level floor encounters rough area where the coefficient of friction is 0.20. The rough area is 5 meters long. How fast will the block be moving as it leaves the rough area? Where did the energy go?**

**2. A 800 kg car traveling southward at 9.2 m/s collides with a 900 kg car traveling northward at 4.5 m/s.?
The cars become entangled. What is the speed and direction of the cars immediately following the collision?**

**3. A 6kg block slides down a 2.0 m long frictionless ramp which is inclined 40 degrees above the ground.?
a) find the height of the top of the ramp and then determine the speed at the bottom of the ramp.**

**b) If the block collides with, and sticks to a 8.2 kg block which had been at rest at the bottom of the ramp, how fast would they now move?**

**c) Determine how much energy the two blocks have after the collision.**

**A****NSWERS**

1. The problem can be solved using work energy theorem

i.e; work done against friction = change in KE

Substitute the values and you will get the answer.

2. This is an example for inelastic collision.

According to the law of conservation of momentum,

total momentum before collision = total momentum after collision

Taking **northwards as positive** and southwards as negative,

Substitute in the formula

Substitute the values of m1, m2, v1 and v2 and you can calculate the value of v.

If v is positive, it will be towards north and if negative it will be towards south.

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