Two balls undergo a perfectly elastic head-on collision, with one ball initially at rest. if the incoming ball has a speed of 20

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3 years ago

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Two balls undergo a perfectly elastic head-on collision, with one ball initially at rest. if the incoming ball has a speed of 20

0 m/s . part a part complete what is the final speed of the incoming ball if it is much more massive than the stationary ball? express your answer using two significant figures. v1 = 200 m/s submitprevious answers correct part b part complete what is the final direction of the incoming ball with respect to the initial direction if it is much more massive than the stationary ball? forward submitprevious answers correct part c part complete what is the final speed of the stationary ball if the incoming ball is much more massive than the stationary ball?.

Physics

2 answers:

melamori03 [73] · Answer 1 · 2022-07-03T14:08:06+03:00

what is the final speed of the incoming ball if it is much more massive than the stationary ball? express your answer using two significant figures. v1 = 200 m / s submitprevious answers correct
Perfectly elastic collisions means that both mechanical energy and
momentum are conserved.
Therefore, for this case, we have the equation to find the final velocity of the incoming ball is given by
v1f = ((m1-m2) / (m1 + m2)) v1i
where,
v1i: initial speed of ball 1.
v1f: final speed of ball 1.
m1: mass of the ball 1
m2: mass of the ball 2
Since the mass of the ball 1 is much larger than the mass of the ball 2 m1 >> m2, then rewriting the equation:
v1f = ((m1) / (m1) v1i
v1f = v1i
v1f = 200 m / s
answer
200 m / s
part b part complete what is the final direction of the incoming ball with respect to the initial direction if it is much more massive than the stationary ball? forward submitprevious answers correct

Using the equation of part a, we can include in it the directions:
v1fx = ((m1-m2) / (m1 + m2)) v1ix
v1i: initial velocity of ball 1 in the direction of the x-axis
v1f: final speed of ball 1 in the direction of the x-axis
like m1 >> m2 then
v1fx = v1ix
v1fx = 200 m / s (positive x direction)
So it is concluded that the ball 1 continues forward.
answer:
forward

part c part complete what is the final speed of the stationary ball if the incoming ball is much more massive than the stationary ball ?.
The shock is perfectly elastic. For this case, we have that the equation to find the final velocity of the stationary ball is given by
v2f = ((2m1) / (m1 + m2)) v1i
where,
v1i: initial speed of ball 1.
v2f: final speed of ball 2.
m1: mass of the ball 1
m2: mass of the ball 2
Then, as we know that m1 >> m2 then
v2f = ((2m1) / (m1) v1i
v2f = 2 * v1i
v2f = 2 * (200 m / s)
v2f = 400 m / s
answer
400m / s

weqwewe [10] · Answer 2 · 2022-07-03T16:27:22+03:00

a) The final speed of the massive incoming ball is $\boxed{200\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}}$ .

b) The direction of the motion of the incoming ball with respect to its initial position is in the forward direction.

c) The final speed of the stationary ball due to the incoming massive ball is $\boxed{{\text{400}}\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}}$ .

Further Explanation:

Given:

The speed of the incoming ball is $200\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ .

The second ball is initially at rest So, its initial velocity is $0\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ .

Concept:

Part (a):

The two bodies moving towards one another when collide elastically; the momentum as well as the mechanical energy of the system remains conserved at the time of collision.

By using the conservation of momentum at the time of collision:

$Mu{ & _1}+m{u_2}=M{v_1}+m{v_2}$ …… (1)

Here, $M$ is the mass of the massive ball, $m$ is the mass of lighter ball, ${u_1}\,\& \,{u_2}$ are the initial speeds of the two bodies and ${v_1}\,\& \,{v_2}$ are the final speeds of the bodies.

Substitute the values of ${u_1}\,\& \,{u_2}$ in above expression.

$M\times200+m\times0=\left({M\times {v_1}} \right)+\left({m\times{v_2}}\right)$

Since $M>> >m$ , the above expression can be rearranged as:

$\begin{aligned}200\times M&=\left(M\right){v_1}\\{v_1}&=200\left({\frac{M}{M}} \right)\\&=200\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}$

Thus, the final speed of the incoming ball at collision is $\boxed{200\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}}$ .

Part (b):

The ball is moving in the forward direction and has no other velocity other than the forward direction. Initially, the ball started in the forward direction and finally it is moving with the speed $200\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ in the forward direction only.

Therefore, it suggests that the massive ball is moving in forward direction with respect to its initial direction.

Part (c):

In order to obtain the final speed of the smaller ball after the collision, rearrange the equation (1) to obtain the final velocity of the second ball ${v_2}$ .

$\begin{aligned}{v_2}&={v_1}\left({\frac{{2M}}{M}}\right)\\&=200\left(2\right)\\&=400\,{{\text{m}} \mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}$

Thus, the final speed of the smaller ball after collision with the massive ball is $\boxed{400\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}}$ in forward direction.