Question

A small explosive charge is placed in a rubber block resting on a smooth surface. When the charge is detonated, the block breaks into three pieces. A 200-g piece travels at 1.4 m/s [N], and a 300-g piece travels at 0.90 m/s E [10°] N? (cont.)

The remaining piece flies off at a speed of 1.8 m/s. Calculate the mass and velocity of the third piece. Assume two significant digits for each value.

Answer 1

`m = 1.59" kg"` `theta = 264^@` This is `6^@` West of South.

Explanation:

Before we begin I will change North to be `90^@` and `10^@` North of East to be `10^@`, because this corresponds with the standard angles for the trigonometric functions.

The momentum vector of the first piece is:

`vec(p_1) = (0.20" kg")(1.40" m/s")((cos(90^@))hati+sin(90^@)hatj)`

`vec(p_1) = (2.80" kg•m/s")hatj`

The momentum vector of the second piece is:

`vec(p_2) = (0.30" kg")(0.90" m/s")((cos(10^@))hati+sin(10^@)hatj)`

`vec(p_2) = (0.27" kg•m/s")((cos(10^@))hati+sin(10^@)hatj)`

`vec(p_2) = (0.26" kg•m/s")hati+(0.05" kg•m/s")hatj`

The momentum vector of the third piece is:

`vec(p_3) = m(1.80" m/s")((cos(theta))hati+sin(theta)hatj)`

`vec(p_3) = m(1.80" m/s")(cos(theta))hati+m(1.80" m/s")sin(theta)hatj`

Summing the vectors in the `hati` direction:

`(0.26" kg•m/s")hati + m(1.80" m/s")(cos(theta))hati = 0`

`mcos(theta) = -(0.26" kg")/1.80" [1]"`

Summing the vector in the `hatj` direction:

`(2.80" kg•m/s")hatj + (0.05" kg•m/s")hatj+ m(1.80" m/s")sin(theta)hatj = 0`

`msin(theta) = -(2.85" kg")/1.80" [2]"`

Divide equation [2] by equation [1]:

`tan(theta) = 2.85/0.26`

Because the sine and the cosine functions are both negative we must adjust the angle for the 3rd quadrant:

`theta = tan^-1(2.85/0.26)+ 180^@`

`theta = 264^@`

This is `6^@` West of South.

We can use either equation [1] or [2] to compute the mass; I will use equation [2]:

`msin(264^@) = -(2.85" kg")/1.80`

`m = -(2.85" kg")/(1.80sin(264^@))`

`m = 1.59" kg"`