Newton's Laws of Motion
Trending Questions
Q. A body of mass m1 collides head on elastically with a stationary body of mass m2. If velocities of m1 before and after the collision are v and –v/3 respectively then the value of m1/m2 is
- 0.5
- 4
- 1
- 2
Q. In the figure shown below, mA=2 kg and mB=3 kg. A force F=40 N is applied on 2 kg block as shown. Find the acceleration of COM of the system. (Neglect friction everywhere and assume the string to be inextensible)
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/843829/original_9.png)
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/843829/original_9.png)
- 1.2 m/s2
- 1 m/s2
- 2.4 m/s2
- 2 m/s2
Q. Two blocks of masses 5 kg and 2 kg are placed at rest on a frictionless surface and connected by a spring. An external hit gives a velocity of 21 m/s to the heavier block towards the lighter one. Velocities of both blocks (heavier and lighter one) in the centre of mass frame, just after the kick will be respectively:
(Consider direction of motion of heavier block as +ve direction)
(Consider direction of motion of heavier block as +ve direction)
- 6 m/s, 10 m/s
- 6 m/s, −15 m/s
- 15 m/s, 15 m/s
- 3 m/s, 5 m/s
Q. For the arrangement shown in the figure, m1=6 kg and m2=4 kg. If the string is light and inextensible and the pulley and surfaces are frictionless, find the magnitude of the acceleration of the center of mass.
(Take g=10 m/s2)
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/843827/original_8.png)
(Take g=10 m/s2)
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/843827/original_8.png)
- 4 m/s2
- 10 m/s2
- 6 m/s2
- 2.88 m/s2
Q. For a spring block system as shown in figure, a time varying force F=5t N is applied on mass 2 kg. After 10 seconds, velocity of 3 kg mass is 30 m/s. Find the velocity of 2 kg mass at this instant.
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/897830/original_4.png)
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/897830/original_4.png)
- 40 m/s
- 80 m/s
- 803 m/s
- 20 m/s
Q. A body of mass 10 kg moving with velocity of 10 m/s hits another body of mass 30 kg moving with velocity 3 m/s in same direction. The co-efficient of restitution is 14. The velocity of centre of mass after collision will be
- 20 m/s
- 40 m/s
- 194m/s
- 234m/s
Q. For a spring block system as shown in figure, a time varying force F=5t N is applied on mass 2 kg. After 10 seconds, velocity of 3 kg mass is 30 m/s. Find the velocity of 2 kg mass at this instant.
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/897830/original_4.png)
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/897830/original_4.png)
- 40 m/s
- 80 m/s
- 803 m/s
- 20 m/s
Q. A man of mass 75 kg is standing on a platform of mass 15 kg kept on a smooth horizontal surface. The man starts moving on the platform with a velocity of 20 m/s relative to the platform. Then, magnitude of the recoil velocity of the platform is
- 12.33 m/s
- 20 m/s
- 25 m/s
- 16.66 m/s
Q. Two blocks of masses 5 kg and 2 kg are placed at rest on a frictionless surface and connected by a spring. An external hit gives a velocity of 21 m/s to the heavier block towards the lighter one. Velocities of both blocks (heavier and lighter one) in the centre of mass frame, just after the kick will be respectively:
(Consider direction of motion of heavier block as +ve direction)
(Consider direction of motion of heavier block as +ve direction)
- 6 m/s, 10 m/s
- 6 m/s, −15 m/s
- 15 m/s, 15 m/s
- 3 m/s, 5 m/s
Q. In the figure shown below, mA=2 kg and mB=3 kg. A force F=40 N is applied on 2 kg block as shown. Find the acceleration of COM of the system. (Neglect friction everywhere and assume the string to be inextensible)
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/843829/original_9.png)
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/843829/original_9.png)
- 1.2 m/s2
- 1 m/s2
- 2.4 m/s2
- 2 m/s2
Q. Two blocks are attached to the pulley as shown in the arrangement.
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/844352/original_16.png)
If the string is inextensible and all surfaces are frictionless, find out the magnitude acceleration of COM of the system. (Take g=10 m/s2)
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/844352/original_16.png)
If the string is inextensible and all surfaces are frictionless, find out the magnitude acceleration of COM of the system. (Take g=10 m/s2)
- 1.6 m/s2
- 0.4 m/s2
- 0.16 m/s2
- 4 m/s2
Q. A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x)= bx−2n, where b and n are constants and x is the position of the particle. The acceleration of the particle as a function of x, is given by
Q. A particle of mass m is moving with a constant acceleration 2a towards another particle of mass 4m as shown in figure. Mass 4m is also moving with acceleration a towards the particle of mass m. Find the acceleration of COM of the system.
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/882155/original_2.png)
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/882155/original_2.png)
- a/5 (towards left)
- 2a/5 (towards left)
- a/5 (towards right)
- 2a/5 (towards right)
Q. A man wearing a bullet-proof vest stands still on roller skates. The total mass is 80 kg. A bullet of mass 20 grams is fired at 400 m/s. It is stopped by the vest and man falls to the ground. What is then, the velocity of man ?
- 1 m/s
- 0.1 m/s
- 2 m/s
- 2.5 m/s
Q. A pulley mass arrangement is shown is the figure. A force of 50 N is applied on the 2 kg block. Find the acceleration of COM of the system.
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/876346/original_1.12.png)
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/876346/original_1.12.png)
- 1.6 m/s2
- 2 m/s2
- 10 m/s2
- 0 m/s2
Q. A man wearing a bullet-proof vest stands still on roller skates. The total mass is 80 kg. A bullet of mass 20 grams is fired at 400 m/s. It is stopped by the vest and man falls to the ground. What is then, the velocity of man ?
- 1 m/s
- 2 m/s
- 2.5 m/s
- 0.1 m/s