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51.

A ping-pong ball strikes a wall with a velocity of `10 ms^(-1)`. If the collision is perfectly elastic, find the velocity if ball after impactA. `-20 ms^(-1)`B. `-5 ms^(-1)`C. `1.0 ms^(-1)`D. `-10 ms^(-1)`

Answer» Correct Answer - D
`m_(1)u_(1) +m_(2) u_(2) = m_(1)v_(1) + m_(2)v_(2)`
Discussion
52.

An electric motor operates with an efficiency of `90%`. A pump operated by the motor has an efficiency of `80%`. The overall efficiency of the system isA. `85%`B. `100%`C. `72%`D. `60%`

Answer» Correct Answer - C
`eta = eta_(1) xx eta_(2)`
Discussion
53.

A body of mass `2kg` is projected vertically up with velocity `5ms^(-1)`. The work done on the body by gravitational force before it is brought to rest momentarily isA. `250 J`B. `25J`C. `0J`D. `-25 J`

Answer» Correct Answer - D
`W = -mgh = -(1)/(2)m u^(2)`
Discussion
54.

A railway truck of mass `16000 kg` moving with a velocity of `5ms^(-1)` strikes another truck of mass `4000 kg` at rest. If they move together after impact, their common velocity isA. `2ms^(-1)`B. `4ms^(-1)`C. `6 ms^(-1)`D. `8ms^(-1)`

Answer» Correct Answer - B
`vec(v) = (m_(1)vec(u_(1))+m_(1)vec(u_(2)))/(m_(1)+m_(2))`
Discussion
55.

Block `A` is hanging from vertical spring of spring constant `K` and is rest. Block `B` strikes block `A` with velocity `v` and sticks to it. Then the value of `v` for which the spring just attains natural length is A. `sqrt((60mg^(2))/(k)`B. `sqrt((6mg^(2))/(k)`C. `sqrt((10mg^(2))/(k)`D. `sqrt((mg^(2))/(k)`

Answer» Correct Answer - B
`x_(0) = (mg)/(k)`, after the collision speed of combined mass is `(V)/(2)`.
`(1)/(2) xx 2m((v^(2))/(2))+(1)/(2)k((mg)/(k))^(2) = 2 mg((mg)/(k))` so, `V = sqrt((6mg^(2))/(k))`
Discussion
56.

Two point masses are connected by a light intextensible string are lying on a frictionless surface as shown in figure. An impulse of magnitude `10 kg-m//s` is given to `5 kg` block. A. Velocity of `10 kg` block immediately after impulse is given `(1)/(3) m//s`B. Velocity of `10 kg` block immediately after impulse is given `2 m//s`C. Speed of `5 kg` block immediately after impulse is given `sqrt((28)/(9)) m//s`D. Speed of `5 kg` block immediately after impulse is given `(sqrt(28))/(9) m//s`

Answer» Correct Answer - A::C
Along the string `10 cos 60^(circ) = (15)v_(x)`
`rArr v_(x) = (1)/(3)m//s` for both `10 kg` and `5 kg`.
In the direction perpendicular to string
`10 sin 60^(circ) = 5v_(y) rArr v_(y) = sqrt(3) m//s`
Velocity of `10 kg = (1)/(3) m//s`,
velocity od `5 kg = sqrt(3+(1)/(9)) = (sqrt(28))/(3)m//s`
Discussion
57.

A small ball falling vertically downward with constant velocity `4m//s` strikes elastically a massive inclined cart moving with velocity `4m//s` horizontally as shown. The velocity of the rebound of the ball is A. `402 m//s`B. `403 m//s`C. `4m//s`D. `405 m//s`

Answer» Correct Answer - C
Discussion
58.

A small ball is projected at an angle `alpha` between two vertical walls such that in the absence of the wall its range would have been `5d`. Given that all the collisions are perfectly elastic, find. (a) maximum height atained by the ball. (b) total number of collisions with the walls before the ball comes back to the ground, and (c) point at which the ball finally falls. The walls are supposed to be very tall. A. `(2u^(2) sin^(2) alpha)/(g)`B. `(2u^(2) cos^(2) alpha)/(g)`C. `(u^(2) sin^(2) alpha)/(2g)`D. `(u^(2))/(2g)`

Answer» Correct Answer - C
Vertical component of velocity of the ball do not change after collision.
Discussion
59.

A block of mass `1 kg` moving with a speed of `4 ms^(-1)`, collides with another block of mass `2 kg` which is at rest. If the lighter block comes to rest after collision, then the speed of the heavier body isA. `2 ms^(-1)`B. `1 ms^(-1)`C. `1.5 ms^(-1)`D. `0.5 ms^(-1)`

Answer» Correct Answer - A
`m_(1)u_(1)+m_(2)u_(2) = m_(1)v_(1)+m_(2)v_(2)`
Discussion
60.

A bullet of mass `50` grams going at a speed of `200 ms^(-1)` strikes a wood block of mass `950 gm` and gets embedded in it. The velocity of the block after the impact isA. `5 ms^(-1)`B. `10 ms^(-1)`C. `20 ms^(-1)`D. `50 ms^(-1)`

Answer» Correct Answer - B
`v = (m_(1)u_(1)+m_(2)u_(2))/(m_(1)+m_(2))`
Discussion
61.

Particles `P` and `Q` of masses `20g` and `40g`, respectively, are projected from positions `A` and `B` on the ground. The initial velocities of `P` and `Q` make angles of `45^(circ)` and `135^(circ)`, respectively with the horizontal as shown in the fig. Each particle has an initial speed of `49m//s`. The separation `AB` is `245m`. Both particles travel in the same vertical plane and undergo a collision. After the collision `P` retraces its path. The separation of `Q` from its initial position when it hits the ground is A. `245m`B. `(245)/(3)m`C. `(245)/(2)m`D. `(245)/(sqrt(2))m`

Answer» Correct Answer - C
Both the particles will collide at the highest point of their path. Momentum of the particle `P` just before collision
`m_(A)V_(A) cos 45^("circ") = 20 xx 10^(-3) xx 49 xx (1)/(sqrt(2))`
`(980)/(sqrt(2)) xx 10^(-3)kg m//s`
Momentum of particle `Q` just before collision is
`m_(B)V_(B) cos 135^("circ") = -40 xx 10^(-3) xx 49 xx (1)/(sqrt(2)) = (1960)/(sqrt(2)) xx 10^(-3) kg m//s`
After collision, let velocity of `Q` be `V` and velocity of `P` be `49 cos 45^("circ")`. Therefore, momentum after collision is
`40 xx 10^(-3)V - (49 xx 20 xx 10^(3))/(sqrt(2))`
Applying law of conservation of momentum
`(980)/(sqrt(2)) xx 10^(-3) - 1960 xx (10^(-3))/(sqrt(2)) = 40 xx 10^(-3)V-(980 xx 10^(-3))/(sqrt(2))`
or `v = 0`
Thus, particle `Q` will fall freely after collision. So, the distance travelled by it will be `245//2m`.
i.e., `122.5m`
Discussion