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1.

Two paricle A and B initially at rest, move towards each other under mutual force of attraction. At the instant when the speed of A is V and the speed of B is 2V, the speed of the centre of mass of the system isA. zeroB. `v`C. `(3v)/(2)`D. `(3y)`

Answer» Correct Answer - A
Initially , `v_(c.m.) = 0`
Since there is no external force , `v_(c.m.)` remains same. Therefore `v_(c.m.)` is always zero.
Discussion
2.

A paticle of mass `m` is executing uniform circular motion on a path of radius `r`. If `p` is the magnitude of its linear momentum, then the radial force acting on the particle isA. `p m r`B. `r m//p`C. `m p^(2)//r`D. `p^(2) //r m`

Answer» Correct Answer - D
`F = (mv^(2))/(r ) = (m)/(r ) ((p)/(m))^(2) = (p^(2))/(m r)`
Discussion
3.

A body falling from a height of `10 m` rebounds from the hard floor . ItA. `0.89`B. `0.56`C. `0.23`D. `0.18`

Answer» Correct Answer - A
`h = 10m , h_(1) = (1 - (20)/(100)) h = 0.8 h`
`h_(1) = e^(2) h rArr e = sqrt(0.8) = 0.89`
Discussion
4.

The rate of mass of the gas emitted from the rear of a rocket is initially `0.1 kg//s`. If the speed of the gas relative to the rocket is `50 m//s` and the mass of the rocket is `2 kg` , then the acceleration of the rocket in `m//s^(2)` isA. `5`B. `5.2`C. `2.5`D. `25`

Answer» Correct Answer - C
Thrust `F = v_(r ) (dm)/(dt)`
`= 50(0.1) = 5`
`a = (F)/(m) = (5)/(2) = 2.5 m//s^(2)`
Discussion
5.

Which of the following is true for center of mass ? (i) The center of mass of a body may lie within , outside , on the surface of the body. (ii) In the case of symmetrical bodies , the center of mass coincides with the geometrical center of the body. (iii) In the absence of external forces , the center of mass moves with constant velocity. (iv) If external forces are absent and system is initially at rest, then location of center of mass is fixed.A. `(i) ,(ii)`B. `(i) ,(ii) ,(ii)`C. `(ii) ,(iii) ,(iv)`D. all options are correct

Answer» Correct Answer - D
Discussion
6.

Two balls are thrown simultaneously from top of tower in air as shown in the figure. (i) The acceleration of the center of mass of two balls while in air is equal to `g`. (ii) The path followed by the center of mass is parabola. (iii) The path followed by the center of mass will change if one ball after striking the ground comes to rest. A. `(i),(ii) ,(iii)`B. `(i),(ii) ,(iv)`C. `(i),(ii)`D. all of the above

Answer» Correct Answer - C
`m_(1) = m_(1) , vec(a)_(1) = -g hat(j) ,m_(2) = 2m , vec(a)_(2) = -g hat(j)`
`vec(a)_(c.m.) = (m_(1) vec(a)_(1) + m_(2) vec(a)_(2))/(m_(1) + m_(2)) = (-m g hat(j))/(3m) = -g hat(j)`
`|vec(a)_(c.m.)| = g , (i) is O.K`.
`(m + 2m) vec(u)_(c.m.) = (mu_(1) cos alpha + 2mu_(2) cos beta) hat(i)`
`+ (mu_(1) sin alpha - 2m u_(2) sin beta) hat(j)`
`= u_(x) hat(i) + u_(y) hat(j)`
`vec(x)_(c.m.) = u_(x) hat(i) t , vec(y)_(c.m.) = u_(y) t hat(j) - (1)/(2) gt^(2) hat(j)`
Trajectory `y = (u_(y))/(u_(x)) x - (1)/(2) (gx^(2))/(u_(x)^(2))`, parabola
Discussion
7.

Two blocks of mass `1kg` and `3 kg` have position v ectors ` hat(i) + 2 hat(j) + hat(k)` and `3 hat(i) - 2 hat(j) + hat(k)` , respectively . The center of mass of this system has a position vector.A. `- 2 hat(i) + 2 hat(k)`B. `-2 hat(i) - hat(j) + hat(k)`C. `2.5 hat(i) - hat(j) - hat(k)`D. `- hat(i) + hat(j) + hat(k)`

Answer» Correct Answer - C
`vec(R )_(c.m.) = (m_(1) vec(R ) + m_(2) vec (R ))/(m_(1) + m_(2))`
`= - 2.5 hat(i) - hat(j) + hat(k)`
Discussion
8.

The force `F` acting on a partical of mass `m` is indicated by the force-time graph shown below. The change in momentum of the particle over time interval from zero to 8 `s` is. A. `20 Ns`B. `12 Ns`C. `6 Ns`D. `24 Ns`

Answer» Correct Answer - B
The area of `F - T` graph gives change in linear momentum.
`t = 0 to t = 8`
`Delta p = (1)/(2) xx 6 xx 2 - 3 xx 2 + 3 xx 4`
`6 - 6 + 12 = 12 Ns`
Discussion
9.

A cubical block of ice of maas m and edge L is placed in a large tray of maas M. If the ie melts, how far does the centre of maas of the system 'ice plus tray' come down?A. `(mL)/((m + M))`B. `(mL)/(2(m + M))`C. `(mL)/(M)`D. `(ML)/(m)`

Answer» Correct Answer - B
`(y_(c.m.))_(1) = (m (L)/(2) + M xx 0)/(m + M) = (m L)/(2( m + M))`
When ice becomes water , as tray is large , the height of water is negligible
`(y_(c.m.))_(2) = 0`
Shift `= (y_(c.m.))_(1) - (y_(c.m.))_(2) = (mL)/(2(m + M))`
Discussion
10.

A ball hits the floor and rebounds after an inelastic collision. In this caseA. `(i) ,(ii)`B. `(i) ,(iv)`C. `(ii) ,(iii)`D. `(iii) ,(iv)`

Answer» Correct Answer - D
Discussion
11.

Look at the drawing given in the figure which has been drawn with ink of uniform line-thickness. The mass of ink used to draw each of the two inner circles, and each of the two line segments is m. The mass of the ink used to draw the outer circle is 6 m. The coordinates of the centres of the different parts are: outer cicle (0,0), left circle (-a, a), right inner circle (a,a), vertical line (0,0) and horizontal line (0 ,-a). The y-coordinate of the centre of mass of the ink in this drawing is A. `(a)/(10)`B. `(a)/(8)`C. `(a)/(12)`D. `(a)/(3)`

Answer» Correct Answer - A
`y_(c.m.) = ( 6 m xx 0 + m xx 0 + m(-a) + ma + ma)/( 6m + m + m + m + m = 10 m) = (a)/(10)`
Discussion
12.

A 500kg rocket is set for verticle firing. The exhaust speed is 800`ms^-2` . To give an initial upward acceleration of 20`ms^-2` , the amount of gas ejected per second to supply the needed thrust will be (g=10 `ms^-2` )A. `127.5 kg//s`B. `187.5 kg//s`C. `85 kg//s`D. `137.5 kg//s`

Answer» Correct Answer - B
`Ma = v_(r ) (dm)/(dt)`
`v_( r) (dM)/(dt) = M(a + g)`
`(800) (dM)/(dt) = 5000 (20 + 10)`
`(dM)/(dt) = 187.5 kg//s`
Discussion
13.

A bomb of mass `9 kg` explodes into `2` pieces of mass `3 kg` and `6 kg`. The velocity of mass `3 kg is 1.6 m//s`. The `K.E. of mass 6 kg` isA. `3.84 J`B. `9.6 J`C. `1.92 J`D. `2.92 J`

Answer» Correct Answer - C
Discussion
14.

An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other . The first part of mass `1 kg` moves with a speed of `12 m//s` and the second part of mass `2 kg` moves with `8 m//s`. If the third part flies off with `4 m//s` the speed , then its mass is

Answer» Correct Answer - B
Discussion
15.

A particle of mass `m` , initially at rest , is acted upon by a variable force `F` for a brief interval of time `T`. It begins to move with a velocity `u` after the forrce stops acting . `F` is shown in the graph as a function of time. The curve is a semicircle. A. `u = (pi F_(0)^(2))/(2m)`B. `u = (pi T^(2))/(8m)`C. `u = (pi F_(0) T)/( 4m)`D. `(F_(0) T)/(2 m)`

Answer» Correct Answer - C
Area of `F-t` graph ` = (pi F_(0) T)/(4)`
(Area of ellipse `= pi ab)`
`Delta p = (pi F_(0) T)/(4)`
`m( u - 0) = (pi F_(0) T)/(4) rArr u = (pi F_(0) T)/(4m)`
Discussion
16.

In a head on elastic collision of two bodies of equal massesA. `(i) ,(ii)`B. `(i) ,(ii) ,(iii)`C. `(i) ,(ii) ,(iv)`D. all options are correct

Answer» Correct Answer - D
Discussion
17.

In elastic collision between spheres `P` and `Q` of equal mass but unequal radii , move along a straight line. Which of the following may be correct after the collisions ? (i) `P` comes to rest and `Q` moves with velocity of `P`. (ii) `P` and `Q` move with equal speeds making an angle of `45^(@)` each with original line of motion. (iii) `P` and `Q` move with unequal speeds , making angles of `30^(@)` and `60^(@)` with the original line of motion , respectively. (iv) `P` comes to rest:A. `(i)` onlyB. `(iv)` onlyC. `(iii)` onlyD. `None

Answer» Correct Answer - C
After collision, spheres will move in the perpendicular direction.
(i) If the mass of colliding bodies is same and collision is elastic.
After the head-on collision,velocities will be exchanged and bodies will move in a single straight line.
(ii) If collision is not head - on , bodies will move in the perpendicular direction.
Discussion
18.

In the previous problem , the maximum loss in `K.E.` will beA. `(1)/(2) mv_(0)^(2)`B. `(1)/(3) mv_(0)^(2)`C. `(2)/(3) mv_(0)^(2)`D. `(3)/(4) mv_(0)^(2)`

Answer» Correct Answer - B
For loss of `K.E`. to be maximum , the collision should be completely inelastic , i.e. the block should stick together and move with same velocity after the collision.
`(Delta K)_(max) = (1)/(2) (m_(1) m_(2))/(m_(1) + m_(2)) (u_(1) - u_(2))^(2)`
`= (1)/(2) ( m xx 2m)/( m + 2m) (v_(0) - 0)^(2)`
`= (1)/(3) mv_(0)^(2)`
Discussion