A thin ring of mass 2kg and radius 0.5 m is rolling without on a horizontal plane with velocity 1m/s. A small ball of mass 0.1kg, moving with velocity 20 m/s in the opposite direction hits the ring at a height of 0.75m and goes vertically up with velocity 10m/s. Immediately after the collision A. the ring has pure rotation about its stationaryB. the ring comes to a complete stopC. friction between the ring and the ground is to the leftD. there is no friction between the ring and the ground

A thin ring of mass 2kg and radius 0.5 m is rolling without on a horiz

1.	A thin ring of mass 2kg and radius 0.5 m is rolling without on a horizontal plane with velocity 1m/s. A small ball of mass 0.1kg, moving with velocity 20 m/s in the opposite direction hits the ring at a height of 0.75m and goes vertically up with velocity 10m/s. Immediately after the collision A. the ring has pure rotation about its stationaryB. the ring comes to a complete stopC. friction between the ring and the ground is to the leftD. there is no friction between the ring and the ground
Answer» Correct Answer - C Friction force on the ring

Discussion

No Comment Found

Related InterviewSolutions

A fly wheel rotating about a fixed axis has a kinetic energy of `360 J`. When its angular speed is `30 rad s^(-1)`. The moment of inertia of the wheel about the axis of rotation isA. `0.6kgm^(-2)`B. `0.15kgm^(-2)`C. `0.8kgm^(-2)`D. `0.75kgm^(-2)`
A constant external torque `tau` acts for a very brief period `/_ ` on a rotating system having moment of inertia `I` thenA. the angular velocity of the system will change by `tau/_ //I`B. the angular velocity of the system will change by `tau/_ //I`C. if the system was initially at rest, it will acquire rotational kinetic enegy `(tau/_ )^(2)//2I`D. the kinetic energy of the system will change by `(tau/_ )^(2)//(2I)`
A disc of mass `m` and radius `R` moves in the `x-y` plane as shown in Fig. The angular momentum of the disc about the origin `O` at the instant shown is A. `-5/2mR^(2)omegahatk`B. `7/3mR^(2)omegahatk`C. `-9/2mR^(2)omegahatk`D. `5/2mR^(2)omegahatk`
If `vec(tau)xxvec(L)=0` for a rigid body, where `vec(tau)=` resultant torque & `vec(L)` =angular momentum about a point and both are non-zero. ThenA. `vec(L)=`constantB. `|vec(L)|`=constantC. `|vec(L)|` will increasesD. `|vec(L)|` may increase
A particle has a linear momentum p and position vector r. the angular momentum of this particle about the origin will not be zero under the conditionsA. p=0B. p is perpendicular to r and does not cross rC. p is anti-parallel to rD. the particle passes through the origin.
The end B of the rod AB which makes angle `theta` with the floor is being pulled with a constant velocity `v_(v)` as shown. The length of the rod is `l`.A. `3/5v_(0)`B. `4/5v_(0)`C. `5/3v_(0)`D. `5/4v_(0)`
End `A` of a rod `AB` is being pulled on the floor with a constant velocity `v_(0)` as shown. Taking the length of the rod as `l`, at an instant when the rod makes an angle `37^@` with the horizontal, calculate the velocity of the `CM` of the rodA. `5/7v_(0)` at `tan^(-1)4/3` below horizontalB. `5/7v_(0)` at `tan^(-1)3/4` below horizontalC. `5/6v_(0)` at `tan^(-1)3/4` below horizontalD. `5/6v_(0)` at `tan^(-1) 4/3` below horizontal
A sphere of mass `M` and radius r shown in the figure on a rough horizontal plane. At some instant it has translational velocity `v_(0)`, and rotational velocity about centre `v_(0)//2r`. The percentage change of translational velocity after the sphere start pure rolling: A. `14.28%`B. `7.14%`C. `21.42%`D. none
A circular platform is mounted on a vertical frictionless axle. Its radius is `r = 2 m` and its moment of inertia`I = 200 kg m^2`. It is initially at rest. A `70 kg` man stands on the edge of the platform and begins to walk along the edge at speed `v_0 = 1 m s^-1` relative to the ground. The angular velocity of the platform is.A. `1.2 rads^(-1)`B. `0.4rads^(-1)`C. `2.0rads^(-1)`D. `0.7rads^(-1)`
In the given figure `F=10N, R=1m`, mass of the body is `2kg` and moment of inertia of the body about an axis passing through `O` and perpendicular to the plane of the body is `4kgm^(2)`. `O` is the centre of mass of the body. If the ground is smooth, what is the total kinetic energy of the body after `2s`?A. `25J`B. `16.67J`C. `50J`D. `37.5J`

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment