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3 years ago

9

A bat hits a moving baseball. If the bat delivers a net eastward impulse of 1.3 N-s and the ball starts with an initial horizont

al velocity of 3.8 m/s to the west and leaves with a 5.4 m/s velocity to the east, what is the mass of the ball (in grams)?

1 answer:

Georgia [21]3 years ago

3 0

Answer:

$141.30grams$

Explanation:

First, denote our known values;

$J=1.3N-s(+,east)\\u=-3.8ms^-^1(-,west)\\v=5.4ms^-^1(+,east)\\m=?$

Mass is impulse divided by change in velocity:

$m=\frac{J}{v-u}\\=\frac{1.3}{5.4--3.8}\\\\=\frac{1.3}{9.2}\\=0.1413Kgs$

Hence, the mass of the ball is 141.30grams

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Jack and Jill are maneuvering a 3300 kg boat near a dock. Initially the boat's position is < 2, 0, 3 > m and its speed is

Paraphin [41]

Answer:

The workdone by Jack is $W_{jack} = -1050J$

The workdone by Jill is $W_{Jill} = 0J$

The final velocity is $v = 1.36 m/s$

Explanation:

From the question we are given that

The mass of the boat is $m_b = 3300kg$

The initial position of the boat is $P_i = (2 \r i + 0 \r j + 3\r k)m$

The Final position of the boat is $P_f = (4\r i + 0 \r j + 2\r k )\ m$

The Force exerted by Jack $\r F = (-420\r i + 0 \r j + 210\r k) \ N$

The Force exerted by Jill $\r F_{Jill} =(180 \r i + 0\r j + 360\r k)$

Now to obtain the displacement made we are to subtract the final position from the initial position

$\r P = P_f - P_i$

$= (4\r i + 0\r j + 2 \r k) - (2\r i + 0\r j + 3\r k )$

$= (2\r i + 0\r j -\r k )m$

Now that we have obtained the displacement we can obtain the Workdone

which is mathematically represented as

$W =\r F * \r P$

The amount of workdone by jack would be

$W_{jack} =\r F * \r P$

$= [(-420\r i +0\r j +210\r k)(2\r i + 0\r j - \r k)]$

$= (-420) (2) + (210)(-1)$

$= -840 - 210$

$=-1050J$

The amount of workdone by Jill would be

$W_{Jill} =\r F * \r P$

$= [(180 \r i + 0\r j + 360\r k)(2\r i +0\r j -\r k)]$

$= (180 )(2) +(360)(-1)$

$=0J$

According to work energy theorem the Workdone is equal to the kinetic energy of the boat

$W = K.E = \frac{1}{2} m *[v^2 - (1.1)^2]$

$-1050 = 0.5*3300 [*v^2- (1.1)^2]$

$-1050 = 1650 [v^2 -1.21]$

$0.6363 = v^2 -1.21$

$v^2 = 0.6363+1.21$

$v^2 =1.846$

$v = 1.36\ m/s$

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