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A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.68 m/s. The block

Usimov [2.4K]

Answer:

a) $v'=-0.227\ m.s^{-1}$

b) $v=1.36\ m.s^{-1}$

Explanation:

Given:

mass of the lighter block, $m'=0.02\kg$

velocity of the lighter block, $u'=0.68\ m.s^{-1}$

mass of the heavier block, $m=0.04\ kg$

velocity of the heavier block, $u=0\ m.s^{-1}$

a)

Using conservation of linear momentum:

$m'.u'+m.u=m'.v'+m.v$

where:

$v'=$ final velocity of the lighter block

$v=$ final velocity of the heavier block

$m'.u'=m'.v'+m.v$

$m'(u'-v')=m.v$ ........................(1)

Since kinetic energy is conserved in elastic collision:

$\frac{1}{2}m'.u'^2=\frac{1}{2}m'.v'^2+\frac{1}{2}m.v^2$

$m'(u'^2-v'^2)=m.v^2$

$m'(u'-v')(u'+v')= m.v^2$

divide the above equation by eq. (1)

$v=u'+v'$ .............................(2)

now we substitute the value of v from eq. (2) in eq. (1)

$m'(u'-v')=m(u'+v')$

$\frac{m'+m}{m'-m} =\frac{u'}{v'}$

$\frac{0.02+0.04}{0.02-0.04} =\frac{0.68}{v'}$

$v'=-0.227\ m.s^{-1}$ (negative sign denotes that the direction is towards left)

b)

now we substitute the value of v' from eq. (2) in eq. (1)

$m'(u'-v+u')=m.v$

$2m'.u'=(m-m')v$

$2\times 0.02\times 0.68=(0.04-0.02)\times v$

$v=1.36\ m.s^{-1}$

6 0

3 years ago

1. A 4000-kg truck traveling with a velocity of 20 m/s due south collides headon with a 1350-kg car traveling with a velocity of

velikii [3]

(a) The momentum of each vehicle prior to collision is 80000 kgm/s for truck and 13500 kgm/s for car.

(b) The size of momentum is 93500 kgm/s and it will be directed towards South.

Explanation:

The mass of the truck moving due south is given as 4000 kg and the speed is 20 m/s. Similarly, the mass of the car moving due north is 1350 kg and the speed is 10 m/s.

(a) Then the momentum of each vehicle can be obtained by the product of mass with their respective speed.

$Momentum of truck = Mass * Speed = 4000 * 20 =80000 kgm/s$

Similarly, the momentum of car will be

$Momentum of car = 1350 * 10 = 13500 kgm/s$

So, the momentum of each vehicle prior to collision is 80000 kgm/s for truck and 13500 kgm/s for car.

(b) Since, after collision, the vehicles stick together, the momentum after collision will be equal to the total momentum of both the vehicles before collision. This is because, it will obey conservation of momentum.

Momentum of vehicles after collision = total momentum before collision

Momentum after collision = 80000+13500 = 93500 kgm/s.

The direction of the vehicles after collision will be towards south as the mass and speed of the truck is greater than car. So the impact or force exerted on the car by the truck will be greater and thus both the vehicles will be directed towards south after collision.

Thus, the size of momentum is 93500 kgm/s and it will be directed towards South.

4 0

3 years ago

As a substance melts, heat energy is used to break the connections between molecules, and temperature __________ until all melti

mario62 [17]

Temperature stays the same

4 0

3 years ago

How far away is Earth away from the sun?

alexira [117]

It is 92.96 millions miles away

Hope that helped :)

3 0