हिंदी

A cricket ball of mass 150 g has an initial velocity u=(3i^+4j^) m s−1 and a final velocity v = -(3i^+4j^) m s−1 after being hit. - Physics

Advertisements
Advertisements

प्रश्न

A cricket ball of mass 150 g has an initial velocity `u = (3hati + 4hatj)` m s−1 and a final velocity `v = - (3hati + 4hatj)` m s−1 after being hit. The change in momentum (final momentum-initial momentum) is (in kg m s1)

विकल्प

  • zero

  • `-(0.45 hati + 0.6 hatj)`

  • `-(0.9 hati + 1.2 hatj)`

  • `-5 (hati + hatj)`

MCQ

उत्तर

`-(0.9 hati + 1.2 hatj)`

Explanation:

Given, `u = (3hati + 4hatj)` m/s

And `v = - (3hati + 4hatj)` m/s

Mass of the ball = 150 g = 0.15 kg

Δp = Change in momentum

= Final momentum – Initial momentum

= `mv - mu`

= `m(v - u) = (0.15)  [- (3hati + 4hatj) - (3hati + 4hatj)]`

= `(0.15) xx [ - 6hati - 8hatj]`

= `- [0.15 xx 6hati + 0.15 xx 8hatj]`

= `- [0.9 hati + 1.20 hatj]`

Hence, Δp = `-[0.9 hati + 1.2 hatj]`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Laws of Motion - Exercises [पृष्ठ ३०]

APPEARS IN

एनसीईआरटी एक्झांप्लर Physics [English] Class 11
अध्याय 5 Laws of Motion
Exercises | Q 5.3 | पृष्ठ ३०

संबंधित प्रश्न

A block of mass m is placed on a smooth wedge of inclination θ. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block has a magnitude.


Suppose the ceiling in the previous problem is that of an elevator which is going up with an acceleration of 2.0 m/s2. Find the elongation.


The force of buoyancy exerted by the atmosphere on a balloon is B in the upward direction and remains constant. The force of air resistance on the balloon acts opposite the direction of velocity and is proportional to it. The balloon carries a mass M and is found to fall to the earth's surface with a constant velocity v. How much mass should be removed from the balloon so that it may rise with a constant velocity v?


Consider the Atwood machine of the previous problem. The larger mass is stopped for a moment, 2.0 s after the system is set into motion. Find the time that elapses before the string is tight again.


A body of mass m moving with a velocity v is acted upon by a force. Write an expression for change in momentum in each of the following cases: (i) When v << c, (ii) When v → c and (iii) When v << c but m does not remain constant. Here, c is the speed of light.


Show that the rate of change of momentum = mass × acceleration. Under what condition does this relation hold?


Two balls A and B of masses m and 2 m are in motion with velocities 2v and v, respectively. Compare:

(i) Their inertia.

(ii) Their momentum.

(iii)  The force needed to stop them in the same time.


A force of 10 N acts on a body of mass 2 kg for 3 s, initially at rest. Calculate : Change in momentum of the body.


A pebble is dropped freely in a well from its top. It takes 20 s for the pebble to reach the water surface in the well. Taking g = 10 m s-2 and speed of sound = 330 m s-1. Find : The depth of water surface


State Newton's second law of motion.
A body of mass 400 g is resting on a frictionless table. Find the acceleration of the body when acted upon by a force of 0.02 N.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×