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1 year ago

Part A

Physics

1 answer:

irina [24]1 year ago

3 0

Answer:

v' = -18 m/s

Explanation:

Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

$p_{o} = p_{f} (1)$

The initial momentum can be expressed as follows (taking as positive the initial direction of the ball):

$m_{b} * v_{b} -M_{c}*V_{c} = m_{b} * 18 m/s + (-M_{c}* 20 m/s) (2)$

The final momentum can be expressed as follows (since we know that v'b is opposite to the initial vb):

$-(m_{b} * v'_{b}) + M_{c}*V'_{c} (3)$

If we assume that Mc >> mb, we can assume that the car doesn't change its speed at all as a result of the collision, so we can replace V'c by Vc in (3).
So, we can write again (3) as follows:

$-(m_{b} * v'_{b}) +(- M_{c}*V_{c}) = -(m_{b} * v'_{b}) + (-M_{c} * 20 m/s) (4)$

Replacing (2) and (4) in (1), we get:

$m_{b} * 18 m/s + (-M_{c}* 20 m/s) = -(m_{b} * v'_{b}) + (-M_{c} * 20 m/s) (5)$

Simplifying, and rearranging, we can solve for v'b, as follows:
$v'_{b} = -18 m/s (6)$ , which is reasonable, because everything happens as if the ball had hit a wall, and the ball simply had inverted its speed after the collision.

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e. t = -1.0

f. t = -0.35

g. Solve for t
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i. s(t) = [x(t)^2 + y(t)^2]^(1/2)

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2 years ago

Read 2 more answers

a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres

shusha [124]

Answer:

Part a) When collision is perfectly inelastic

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b) When collision is perfectly elastic

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

Explanation:

Part a)

As we know that collision is perfectly inelastic

so here we will have

$mv_m = (m + M)v$

so we have

$v = \frac{mv_m}{m + M}$

now we know that in order to complete the circle we will have

$v = \sqrt{5Rg}$

$\frac{mv_m}{m + M} = \sqrt{5Rg}$

now we have

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b)

Now we know that collision is perfectly elastic

so we will have

$v = \frac{2mv_m}{m + M}$

now we have

$\sqrt{5Rg} = \frac{2mv_m}{m + M}$

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

6 0

2 years ago

A standing wave of the third overtone is induced in a stopped pipe, 2.5 m long. The speed of sound is The frequency of the sound

NemiM [27]

Answer:

f3 = 102 Hz

Explanation:

To find the frequency of the sound produced by the pipe you use the following formula:

$f_n=\frac{nv_s}{4L}$

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vs: speed of sound = 340 m/s

L: length of the pipe = 2.5 m

You replace the values of n, L and vs in order to calculate the frequency:

$f_{3}=\frac{(3)(340m/s)}{4(2.5m)}=102\ Hz$

hence, the frequency of the third overtone is 102 Hz

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2 years ago

1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n

sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

$(1).\: x =v_0t$

$(2).\: y= 15m-\dfrac{1}{2}gt^2$

where $v_0$ is the initial velocity.

(a).

When the projectile hits the 50m mark, $y=0$ ; therefore,

$0=15-\dfrac{1}{2}gt^2$

solving for $t$ we get:

$t= 1.75s.$

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

$50m = v_0(1.75s)$

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$\boxed{v_0 = 28.58m/s.}$

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

$v_x = 28.58m/s.$