A quantity y is to be determined from the equation y=(px)/q^2

A 0.311 kg tennis racket moving 30.3 m/s east makes an elastic collision with a 0.0570 kg ball moving 19.2 m/s west. Find the ve

Troyanec [42]

The velocity of tennis racket after collision is 14.96m/s

<u>Explanation:</u>

Given-

Mass, m = 0.311kg

u1 = 30.3m/s

m2 = 0.057kg

u2 = 19.2m/s

Since m2 is moving in opposite direction, u2 = -19.2m/s

Velocity of m1 after collision = ?

Let the velocity of m1 after collision be v

After collision the momentum is conserved.

Therefore,

m1u1 - m2u2 = m1v1 + m2v2

$v1 = (\frac{m1-m2}{m1+m2})u1 + (\frac{2m2}{m1+m2})u2$

$v1 = (\frac{0.311-0.057}{0.311+0.057})30.3 + (\frac{2 X 0.057}{0.311 + 0.057}) X-19.2\\\\v1 = (\frac{0.254}{0.368} )30.3 + (\frac{0.114}{0.368}) X -19.2\\ \\v1 = 20.91 - 5.95\\\\v1 = 14.96$

Therefore, the velocity of tennis racket after collision is 14.96m/s

7 0

3 years ago

you have been called to testify as a as an expert witness in a trial involving a head-on collision Car A weighs 1515 pounds and

GrogVix [38]

Answer:

70.6 mph

Explanation:

Car A mass= 1515 lb

Car B mass=1125 lb

Speed of car B is 46 miles/h

Distance before locking, d=19.5 ft

Coefficient of kinetic friction is 0.75

Initial momentum of car B=mv where m is mass and v is velocity in ft/s

46 mph*1.46667=67.4666668 ft/s

$Momentum_B=1125*67.4666668 ft/s$

Initial momentum of car A is given by

$Momentum_A=1515v_a$ where $v_a$ is velocity of A

Taking East as positive and west as negative then the sum of initial momentum is

$1515v_a-(1125*67.4666668 ft/s)$

The common velocity is represented as $v_c$ hence after collision, the final momentum is

$Momentum_final=(m_a+m_b)v_c=(1515+1125)v_c=2640v_c$

From the law of conservation of linear momentum, sum of initial and final momentum equals each other hence

$1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

The acceleration of two cars $a=-\mug=-0.75*32.17=-24.1275 ft/s^{2}$

From kinematic equation

$v^{2}=u^{2}+2as$ hence

$v^{2}-u^{2}=2as$

$0^{2}-(v_c)^{2}=2*-24.1275*19.5$

$v_c=\sqrt{2*24.1275*19.5}=30.67 ft/s$

Substituting the value of $v_c$ in equation $1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

$1515v_a-(1125*67.4666668 ft/s)= 2640*30.67$

$1515v_a=(1125*67.4666668 ft/s)+2640*30.67$

$v_a=\frac {156868.8}{1515}=103.5438 ft/s$

$\frac {103.5438}{1.46667}=70.59787 mph\approx 70.60 mph$

3 0

3 years ago

I NEED HELP ASAP

34kurt

I think it’s gonna be b.

4 0

3 years ago

Need help solving this question.

MatroZZZ [7]

Answer:

See the answers below.

Explanation:

to solve this problem we must make a free body diagram, with the forces acting on the metal rod.

i)

The center of gravity of the rod is concentrated in half the distance, that is, from the end of the bar to the center there is 40 [cm]. This can be seen in the attached free body diagram.

We have only two equilibrium equations, a summation of forces on the Y-axis equal to zero, and a summation of moments on any point equal to zero.

For the summation of forces we will take the forces upwards as positive and the negative forces downwards.

ΣF = 0

$-15+T-W=0\\T-W=15$

Now we perform a sum of moments equal to zero around the point of attachment of the string with the metal bar. Let's take as a positive the moment of the force that rotates the metal bar counterclockwise.

ii) In the free body diagram we can see that the force acts at 18 [cm] of the string.

ΣM = 0

$(15*9) - (18*W) = 0\\135 = 18*W\\W = 7.5 [N]$