Advertisements
Advertisements
प्रश्न
A car is moving with a uniform velocity 30 ms-1. It is stopped in 2 s by applying a force of 1500 N through its brakes. Calculate the following values : The change in momentum of car.
Advertisements
उत्तर
Initial velocity, u = 30 m/s
Final velocity, v = 0
Time, t = 2s
Force, F = 1500 N
Here, a = (v - u)/t = (0 - 30)/ 2 = - 15 ms-2. Here, negative sign indicates retardation.
Now, F = ma.
Or, m = F/a = (1500/ 15) = 100 kg.
Change in momentum = Final momentum - Initial momentum
Or, Δp = m (v - u)
Or, Δp = 100 (0 - 30)
Or, Δp = 3000 kg m/s-1
APPEARS IN
संबंधित प्रश्न
A stone of mass m tied to the end of a string revolves in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are: [Choose the correct alternative]
| Lowest Point | Highest Point | |
| a) | mg – T1 | mg + T2 |
| b) | mg + T1 | mg – T2 |
| c) | `mg + T1 –(m_v_1^2)/R` | mg – T2 + (`mv_1^2`)/R |
| d) | `mg – T1 – (mv)/R` | mg + T2 + (mv_1^2)/R |
T1 and v1 denote the tension and speed at the lowest point. T2 and v2 denote corresponding values at the highest point.
A monkey of mass 40 kg climbs on a rope in given Figure which can stand a maximum tension of 600 N. In which of the following cases will the rope break: the monkey
(a) climbs up with an acceleration of 6 m s–2
(b) climbs down with an acceleration of 4 m s–2
(c) climbs up with a uniform speed of 5 m s–1
(d) falls down the rope nearly freely under gravity?
(Ignore the mass of the rope).

Suppose the ceiling in the previous problem is that of an elevator which is going up with an acceleration of 2.0 m/s2. Find the elongation.
The force of buoyancy exerted by the atmosphere on a balloon is B in the upward direction and remains constant. The force of air resistance on the balloon acts opposite the direction of velocity and is proportional to it. The balloon carries a mass M and is found to fall to the earth's surface with a constant velocity v. How much mass should be removed from the balloon so that it may rise with a constant velocity v?
In a simple Atwood machine, two unequal masses m1 and m2 are connected by a string going over a clamped light smooth pulley. In a typical arrangement (In the following figure), m1 = 300 g and m2 = 600 g. The system is released from rest. (a) Find the distance travelled by the first block in the first two seconds; (b) find the tension in the string; (c) find the force exerted by the clamp on the pulley.

A monkey is climbing on a rope that goes over a smooth light pulley and supports a block of equal mass at the other end in the following figure. Show that whatever force the monkey exerts on the rope, the monkey and the block move in the same direction with equal acceleration. If initially both were at rest, their separation will not change as time passes.

A block A can slide on a frictionless incline of angle θ and length l, kept inside an elevator going up with uniform velocity v in the following figure. Find the time taken by the block to slide down the length of the incline if it is released from the top of the incline.

Define Newton’s second law of motion.
Why is it advantageous to turn before taking a long jump?
Why is catching a slow-moving ball easier than catching a fast-moving ball?
