All categories

3 years ago

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

Physics

1 answer:

arlik [135]3 years ago

3 0

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

You might be interested in

which of the following statements is not true Negatively charged objects attract other negatively charged objects. Positively ch

maxonik [38]

Answer:

negatively charged object attract other negatively objects

Explanation:

opposites attract

3 0

4 years ago

Read 2 more answers

A circular saw spins at 6000 rpm , and its electronic brake is supposed to stop it in less than 2 s. As a quality-control specia

Alona [7]

Answer:

a. yes

Explanation:

The initial speed of the circular saw is:

$\dot n_{o} = \frac{6000}{60} \,\frac{rev}{s}$

$\dot n_{o} = 100\,\frac{rev}{s}$

Deceleration rate needed to stop the circular saw is:

$\ddot n = -\frac{100\,\frac{rev}{s} }{2\,s}$

$\ddot n = - 50\,\frac{rev}{s^{2}}$

The number of turns associated with such deceleration rate is:

$\Delta n = \frac{\dot n^{2}}{2\cdot \ddot n}$

$\Delta n = \frac{\left(100\,\frac{rev}{s} \right)^{2}}{2\cdot \left(50\,\frac{rev}{s^{2}} \right)}$

$\Delta n = 100\,rev$

Since the measured number of revolutions is lesser than calculated number of revolution, the circular saw meets specifications.

8 0

3 years ago

two objects were lifted by machine one object had a massive 2 kg it was lifted at a speed of 2 m/s the other had a massive 4 kg

rosijanka [135]

Kinetic energy =(1/2) (mass) (speed²)

First object: KE = (1/2) (2 kg) (2m/s)² = 4 joules during the lift.

Second object: KE = (1/2) (4kg) (3 m/s)² = 18 joules during the lift.

The second object has more kinetic energy while it's being lifted
than the first object has while it's being lifted. Once they reach their
final heights and stop, neither object has any kinetic energy.

8 0

3 years ago

Can someone please solve this

Zina [86]

Answer:

The gravity is pulling the diver downwards but the rotation of the body means gravity cant pull him down as quickly

Explanation:

8 0

3 years ago

A car is going 3 m/s, then it speeds up for 4 seconds. If it has an acceleration of 5m/s^2, find its final velocity. Which equat

andreyandreev [35.5K]

Well, since the question GIVES you the initial velocity, the acceleration, and the time, and ASKS for the final velocity, you'd be smart to find an equation that USES the initial velocity, the acceleration, and the time, and FINDS the final velocity.

Have a look at equation B ..... vf = vi + a*t . That's pretty durn close !

vf = (initial velocity) + (acceleration)*(time)

vf = (3 m/s) + (5 m/s²)*(4 sec)

vf = (3 m/s) + (20 m/s)

vf = 23 m/s

4 0

3 years ago