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allsm [11]
3 years ago
11

2 objects have a total momentum of 400kg m/s, they collide. Object A’s mass is5kg & object B’s mass is 11kg. After the colli

sion Object B is moving at 15m/s.What is the velocity of Object A AFTER the collision?
Physics
1 answer:
ss7ja [257]3 years ago
3 0

Answer:

Explanation:

We shall apply law of conservation of momentum .

Momentum before collision = momentum after collision .

Momentum before collision = 400 kg m/s

Momentum after collision = 5  x v + 11 x 15

where v is velocity of A after the collision .

5  x v + 11 x 15 = 400

5 v = 400 - 165

5v = 235

v = 47 m /s .

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ivann1987 [24]

Answer:

The other angle is 120°.

Explanation:

Given that,

Angle = 60

Speed = 5.0

We need to calculate the  range

Using formula of range

R=\dfrac{v^2\sin(2\theta)}{g}...(I)

The range for the other angle is

R=\dfrac{v^2\sin(2(\alpha-\theta))}{g}....(II)

Here, distance and speed are same

On comparing both range

\dfrac{v^2\sin(2\theta)}{g}=\dfrac{v^2\sin(2(\alpha-\theta))}{g}

\sin(2\theta)=\sin(2\times(\alpha-\theta))

\sin120=\sin2(\alpha-60)

120=2\alpha-120

\alpha=\dfrac{120+120}{2}

\alpha=120^{\circ}

Hence, The other angle is 120°

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4 years ago
How is a projectile different from an object in free fall
aksik [14]

Answer:

Explanation:

Projectile Motion. Projectile motion is different than free fall: it involves two dimensions instead of one. ... Balls traveling in two dimensions, only one of which experiences acceleration, require two sets of equations: one set for the x-direction and the other for the y-direction.

7 0
3 years ago
Starting from rest, a particle that is confined to move along a straight line is accelerated at a rate of 5.0 m/s2. Which one of
Elanso [62]

Answer:

c) The slope is not constant and increases with increasing time.

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We can take derivative of this with respect to time t to get the equation of slope:

s' = (2.5t^2)' = 2*2.5t = 5t

As time t increase, the slope would increases with time as well.

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Delvig [45]

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