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351.

The position of centre of mass of system of particles at any moment does not depend on.A. masses of the particlesB. internal forces on the particlesC. position of the particlesD. relative distance between the particles

Answer» Correct Answer - B
It does not depend on the internal forces acting on the particle.
Discussion
352.

A large number of particles are placed around the origin, each at a distance R from the origin. The distance of the center of mass of the system from the origin isA. equal to RB. less than or equal to RC.D. greater than R

Answer» Correct Answer - B
As large number of particles is situtated at a distance R from the origin. If particles are uniformly distributed and make a circulalr boundary around the origin, then center of mass will be at the origin.
While, if the particles are not uniformly distributed, then center of mass will lie between particle and origin. This implies that the distnace between center of mass and origin is always less than or equal to R.
Discussion
353.

The position of center of mass of a system of particles does not depend upon theA. mass of particlesB. symmetry of the bodyC. position of the particlesD. nature of particles.

Answer» The position of center of mass of a system of particles does not depend upon the nature of particles.
`r_(cm)` = `(summ_(i)r_(i))/(summ_(i))`
Discussion
354.

Three interacting particles of masses `100g, 200g` and `400g` each hace a velocity of `20 m//s` magnitude along the positive direction of x-axis, y-axis and z-axis. Due to force of interaction the third particle stops moving. The velocity of the second particle is `(10hat(j) + 5hat(k))`. What is the velocity of the first particle ?A. `20hat(i) + 20hat(j) + 70hat(k)`B. `10hat(i) + 20hat(j) + 8hat(k)`C. `30hat(i) + 10hat(j) + 7hat(k)`D. `15hat(i) + 5hat(j) + 60hat(k)`

Answer» Correct Answer - (A)
Initial momentum `= m_(1) vec(v)_(1) + m_(2)vec(v)_(2) + m_(3)vec(v)_(3) = 2hat(i) + 4hat(j) + 8hat(k)`
When the third particle stops the final momentum `= m_(1)vec(v)_(1) + m_(2)vec(v)_(1) + m_(2)vec(v)_(2) + m_(3)vec(v)_(3) = 0.1vec(v)_(1) + 0.2(10hat(j) + 5hat(k)) + 0`
By principle of conservation of momentum `0.1 vec(v)_(1) + 2hat(j) + hat(k) = 2hat(i) + 4hat(j) + 8hat(k), vec(v)_(1) = 20hat(i) + 20hat(j) + 70hat(k)`
Discussion
355.

consider the situastion of the previous problem. Suppose the block of mass `m_1` is pulled by a constant force `F_1` and the other block is pulled by a constant force `F_2`. Find the maximum elongation that the spring will suffer.

Answer» Correct Answer - A::B
Acceleration of mass `m_1=(F_1-|F_2)/(m_1+m-2)`
Similarly acceleration of mass ltbr. `m_2=(F_2-F_1)/(m_1+m_2)`
due to `F_1 and F_2` block of mass `m_1 and m_2` will expereince different acceleration and experience an ineertia force
`:.` Net forc eon `m_1=F_1-m_1a`
`=F_1-m_1xx((F_1-F_2))/(m-1+m-2)`
`(m_1F_1+m_2F_1-m_1F_1+_2m-1)/(m-1+m_2)`
`(m_2F_1+m-1F_2)/(m_1+m_2)`
Similarly net force on `m_2`
`=F_2-m_2xx((F_2-F_1))/(m_1+m_2)`
`=(m_1F_2+m_2F_2-m_2F_2+m_2F_1)/(m_1+m_2)`
`=(m_1.F_2+m_2.F_2)/(m_1+m_2)`
`:.If m_1` is dispaced by a distance `x_1 and x_2 by m_2` the maximum extension of the spring is `x_1+x_2`.
`:.` work done by the bocks =energy stored in the spring.
`rarr (m_2.F_1+m_1F_2)/(m_1+m_2)x(x_1)+(m_2F_1+m_1F_2)/(m_1+m_2)xx_2`
`=(1/2)K(x_1+x_2)^2`
`rarr x_1+x_2=2/k (m_2F_1+m_1F_2)/(m_1+m_2)`
Discussion
356.

A ball of mass 50 g moving at a speed of 2.0 m/s strikes a placne surface at an angle of incidence `45^0`. The ball is reflected by the plane at equal angle of reflection with the same speed. Calculate a. the magnitude of the change in momentum of the ball b. the change in the magnitude of the mometum of the ball.

Answer» `Mass=50g=0.05kg`
`v=(2cos 45^0)veci-2sin45^0vecj`
`v_1=-2cos45^0veci-2sin45^0vecj`
a. Change in momentum `=mvecv-mvecv_1`
`0.05(2cos45^0veci-2sin45^0vecj)-`
`0.05-(-2cos45^0veci-2sin45^0vecj)`
`=0.1cos45^0vec-.1sin45^0veci+`
`0.1cos45^0veci+` ,Brgt `0.1cos45^0veci+0.1sin45^0vecj` ltbr. `=0.2cos45^0vecj`
`:. Magnitude =0.2/sqrt2=0.2828/2`
b The chasnge in magnitude of the momentum of the ball
`|vecP_2|-|vecP_1|-2xx0.5-2xx0.5=0`
Discussion
357.

A ball of mass 50 g moving at a speed of 2.0 m/s strikes a plane surface at an angle of incidence `45^0`. The ball is reflected by the plane at an equal angle of reflection with the same speed. Calculate (a). the magnitude of the change in momentum of the ball (b). the change in the magnitude of the mometum of the ball.

Answer» Correct Answer - A::B::D
(a) `|Deltap|=sqrt(p_1^2+p_2^2-2p_1p_2cos 90^@)`
`=sqrt(p_1^2+p_2^2)`
`=sqrt((0.05xx2)^2+(0.05xx2)^2)`
`=0.14kg-m//s`
(b) Initial and final velocity of wall is zero. Therefore change in momentum is zero.
Discussion
358.

A mass of 10 gm moving with a velocity of 100 cm / s strikes a pendulum bob of mass 10 gm . The two masses stick together. The maximum height reached by the system now is `(g=10m//s^(2))`A. 7.5 cmB. 5cmC. 2.5 cmD. 1.25cm

Answer» Correct Answer - D
From conservation of linear momentum, velocity of combined mass just after collision will be `50 cms^(-1)` as mass has doubled.
Now, `H= (u^(2))/(2g) = ((0.5)^(2))/(20)m = 1.25 cm`
Discussion
359.

A pendulum bob of mass `10^(-2) kg` is raised to a height `5 xx 10^(-2)` m and then released. At the bottom of its swing, it picks up a mass `10^(-3) kg`. To what height will the combined mass rise?

Answer» Velocity of pendulam bob in mean position,
`v_(1) = sqrt(2gh)= sqrt(2 xx 10xx5xx10^(-2)= 1 ms^(-1)`
When the bob picks up a mass `10^(-3)` kg at the bottom, then by conservation of linear momentum, the velocity of coalesced mass is given by
`m_(1)v_(1) + m_(2)v_(2) = (m_(1) + m_(2))/v`
`10^(-2) + 10^(-3) xx 0 = (10^(-2) + 10^(-3))v`
or `v=(10^(-2)/(1.1 xx 10^(-2))` = 10/11 ms^(-1)`
Now, `h=(v^(2)/(2g)) = (10//11)^(2)/(2 xx 10) = 4.1 xx 10^(-2)m`
Discussion
360.

A mass of 10 gm moving with a velocity of 100 cm / s strikes a pendulum bob of mass 10 gm . The two masses stick together. The maximum height reached by the system now is `(g=10m//s^(2))`A. ZeroB. 1.25 cmC. 2.5 cmD. 5cm

Answer» Correct Answer - B
`mu=(m+m)v rArr v=u/2 = 1/2m//s`
n=v^(2)/(2g) = 1/(8g) xx10 = 5/4=1.25`
Discussion
361.

Find the ratio of the linear momenta of two particles of masses 1.0 kg and 4.0 kg if their kinetic energies are equal.

Answer» Correct Answer - A::B
`p=sqrt(2km)`
or `ppropsqrtm`
`:. p_1/p_2=sqrt(m_1/m_2)=sqrt(1/4)=1/2`
Discussion
362.

Three particles of masses `20g`, `30g` and `40g` are initially moving along the positive direction of the three coordinate axes respectively with the same velocity of `20cm//s`. When due to their mutual interaction, the first particle comes to rest, the second acquires a velocity `(10hati+20hatk)cm//s`. What is then the velocity of the third particle?A. `2hati+3hatj+4hatk`B. `2.5hati+15hatj+5hatk`C. `2.5hati+10hatj-5hatk`D. `2hati-3hatj-4hatk`

Answer» Correct Answer - B
`P_(1)=20xx20hati, P_(3)=40xx20hatk`
`P_(2)=30xx20 hatj`
`P_(1)=P_(1)+P_(2)+P_(3)=P_(f)`
`rArr 400i+600hatj+800hatk=30(10i+20 hatk)+40 v`
get `V=(100i+600hatj+200hatk)/40`
`=2.5 hati+15 hatj +5 hatk`
Discussion
363.

Three particles of masses `20g`, `30g` and `40g` are initially moving along the positive direction of the three coordinate axes respectively with the same velocity of `20cm//s`. When due to their mutual interaction, the first particle comes to rest, the second acquires a velocity `(10hati+20hatk)cm//s`. What is then the velocity of the third particle?

Answer» Correct Answer - A::B::C
`p_i=p_f`
`:. (20)(20hati)+30(20hatj)+40(20hatk)`
`=(20)(0)+30(10hati+20hatk)+40v`
Solving this equation we get,
`v=(2.5hati+15hatj+5hatk)cm//s`
Discussion
364.

A particle of mass m, collides with another stationary particle of mass M. If the particle m stops just after collision, then the coefficient of restitution for collision is equal toA. 1B. `m/M`C. `(M-m)/(M+m)`D. `m/(M+m)`

Answer» Correct Answer - B
As net horizontal force acting on the system is zero, hence, momentum must remain conserved.
Hence, `mu + 0 = 0 +Mv_(2)`
As per definition, `e = |(v_(2)-v_(1))|/|(u_(2)-u_(1))| = |(v_(2)-0)/(0-u)|`
`v_(2)/u= (mu)/(M)/u = m/M`
Discussion
365.

A particle of mass `m` moving with a velocity `u` makes an elastic one-dimensional collision with a stationary particle of mass `m` establishing a contact with it for extermely small time. `T`. Their force of contact increases from zero to `F_0` linearly in time `T//4`, remains constant for a further time `T//2` and decreases linearly from `F_0` to zero in further time `T//4` as shown. The magnitude possessed by `F_0` is..A. `("mu")/T`B. `(2"mu")/T`C. `(4"mu")/(3T)`D. `(3"mu")/(4T)`

Answer» Correct Answer - C
Impulse `=` area of tapezium,
`=1/2(T+T/2)F_(0)=(3TF_(0))/4`
According to impulse momentum theorem.
Impulse change in momentum
`implies (3TF_(0))/4="mu"impliesF_(0)=(4"mu")/(3T)`
Discussion
366.

A body `X` with a momentum `p` collides with another identical stationary body `Y` one dimensionally. During the collision, `Y` gives an impulse `J` to body `X`. Then coefficient of restitution isA. `(2J)/p-1`B. `J/P+1`C. `J/P-1`D. `J/(2P)-1`

Answer» Correct Answer - A
`u_(1)=p/m, u_(2)=0, c_(1)=((p-J))/m,v_(2)=J/m`
Now apply `e=(v_(2)-v_(1))/(u_(1)-u_(2))`
Discussion
367.

In a one dimensional collision between two identical particles A and B, B is stationary and A has momentum p before impact. During impact, B given J to A.A. The total momentum of the `A+B` System is `p` before and after impact, and `(p - 1)` during the impact.B. During the impact A given impulse J to BC. the coefficient of restituation is `(2J)/(p) - 1`D. The coefficient of restituation is `(J)/(p) + 1`

Answer» Correct Answer - B::C
`underset(A)overset(P)overset(rarr)"O" + underset(B)"O" = underset(A)"O" larrJ" " Jrarrunderset(B)"O"`
For `A : P - J = mv_(1)` …(i)
for `B : J = mv_(2)` ….(ii)
`:. e = -((v_(1) - v_(1))/(u_(2) - u_(2))) = -[(J)/(m)- (((P -J)/(m)))/(0 - (P)/(m))] = (2J)/(P) -1`
Discussion
368.

A uranium 238 nucleus, initially at rest emits n alpha particle with a speed of `1.4xx10^7m/s`. Calculte the rcoil speed of the residual nucleus thorium 234. Assume that the mas of a nucleus is proprotional to the mass number.

Answer» As uranium 238 nucleus emits alpha-particel with a speed of `1.4xx10^7` m/sec.
Let `v_2` be th speed of the residual nucleus thorium 234.
`m_1v_1=m_2v_2`
`rarr 4xx1.4xx10^7=234xxv_2`
`v_2=(4xx1.4xx10^7)/235`
`=2.4xx10^5 m/sec`
Discussion