302 - Mechanics part 1 Questions Answers

A block of mass m is at rest on a rough inclined plane of angle of inclination θ . If coefficient of friction between the block and the inclined plane is required to move the block on the plane is

how the answer is   mg[sinθ - μ cosθ]  .

 

Asked By: AVANTIKA PURI Solved By: SARIKA SHARMA
is this question helpfull: 1 0 read solutions ( 2 ) | submit your answer
Joshi sir comment

i think one line is missing in the question.

a small body starts sliding from a height h down an inclined groove passing into a half circle of radius h/2. find the speed of the bodywhen it reaches the highest point of its trajectory

Asked By: DANI
is this question helpfull: 2 1 submit your answer
Solution by Joshi sir

since body starts from height h at rest and ends in the same height so it will again be in rest 

BLOCK OF WOOD IS KEPT ON THE FLOOR OF A STATIONARY ELEVATOR. D ELEVATOR BEGINS TO DESCEND WITH AN ACCLN 12m/s2 . IF g =10  , THEN DISPLACEMENT OF BLOCK W.R.T. GROUND DURING FIRST 0.2 SEC. AFTER D START WILL BE             ??

ANS 0.2m

Asked By: SARIKA SHARMA
is this question helpfull: 6 0 submit your answer
Joshi sir comment

s = ut + 1/2 at2

so s = 0 + 1/2 10(0.2)2

so s = 5(0.04) = 0.2

since elevator is moving with an acc. greater than g so we should take it a free fall

A body of mass 32kg is suspended by a spring balance from the roof of a vertically operating lift going downward from rest . At that instant  the lift covered 20m and 50m , the spring balance showed 30 kg & 36 kg respt. the vel. is increasing at 20m & decreasing at 50m. Plzz sir explain this.

Asked By: SARIKA SHARMA
is this question helpfull: 3 1 submit your answer
Joshi sir comment

for downward journey

first case

32g-30g = 2g = 32a so a = g/16

second case

32g-36g = -4g = 32a so a = -g/8

so it is acceleration upto 20 m and retardation after that

 

Physics >> Mechanics part 1 >> Kinematics Medical Exam

d benches of a gallery in a cricket stadium r 1m wide & 1m high . a boy strikes d ball at a level of 1m above d ground & hits a sixer. d ball starts at 35m/s at an < of 530 with d horizontal . d benches r perpendicular to plane of motion & d first bench is 110m from d boy. ON WHICH BENCH WILL THE BALL HIT??

ANS 6TH

Asked By: SARIKA SHARMA
is this question helpfull: 3 0 submit your answer
Joshi sir comment

 

we know that y = xtanθ - gx2/2u2cos2θ

according to the given conditions 

x  = 110+n

y = n                              here n is the number of bench in which ball strikes

θ, u and g are given 

 

A 6 kg bomb at rest explodes into three equal pieces P, Q and R. If P flies with speed 30 m/s and Q with speed 40 m/s making an angle 90° with the direction of P. The angle between the direction of motion of P and R is about

1.    143°        2.   127°     3.  120°

Asked By: AVANTIKA PURI
is this question helpfull: 5 1 read solutions ( 1 ) | submit your answer
Joshi sir comment

since initially bomb is in rest so initially momentum of the bomb = 0i+0j+0k

after explosion total momentum = 30mi+40mj+mv                    here v is a vector

now by momentum conservation 0 = 30mi+40mj+mv so v = -30i-40j

so vector along P is i and along R is -3i-4j

so by A.B = |A||B|cosθ

we get i.[-3i-4j] = 1*5*cosθ

so -3/5 = cosθ so θ = 127

If n balls hit elastically and normally on a surface per unit time and all the balls of mass m are moving with same velocity u, then force on surface is

2 mun . How ?

Asked By: AVANTIKA PURI
is this question helpfull: 2 2 submit your answer
Joshi sir comment

momentum change in elastic collision = mu-(-mu) = 2mu                      here after collision partical will retrace its direction 

for n balls it will be  2mun

so force = rate of change of momentum = 2mun/1 = 2mun

A man of mass 50 kg  carries a bag of weight 40 N on his shoulder.  The force with which the floor pushes up his feet will be

1.  882 N         2.   530 N          3.  90 N         4.   600 N

Asked By: AVANTIKA PURI
is this question helpfull: 5 1 submit your answer
Joshi sir comment

reaction by ground is due to both bag and man so 50g+40 = 50*9.8+40 = 490+40 = 530 N

When a 4 kg rifle is fired, the 10 g bullet receives an acceleration of 3 x 10^6 cm/s^2 . The magnitude of the force acting on the rifle (in newton) is

1. 300      2.  200     3.  3000

Asked By: AVANTIKA PURI
is this question helpfull: 2 0 submit your answer
Joshi sir comment

force on the bullet = [10/1000]*3*10= 300 N

so that will be same for rifle

A body of mass 2 kg is sliding with a constant velocity of 4 m/s on a frictionless horizontal table. The force required to keep the body moving with the same velocity is

1.   8 N      2.   0 N      3. 1/2   N

Asked By: AVANTIKA PURI
is this question helpfull: 3 1 submit your answer
Joshi sir comment

no force is required

Login Here

Register