A small dart of mass 0.020 kg is launched at an angle of 30°...

A small dart of mass 0.020 kg is launched at an angle of 30° above the horizontal with an initial speed of 10 m/s. At the moment it reaches the highest point in its path and is moving horizontally, it collides with and sticks to a wooden block of mass 0.10 kg that is suspended at the end of a massless string. The center of mass of the block is 1.2 m below the pivot point of the string. The block and dart then swing up until the string makes an angle ? with the vertical, as shown above. Air resistance is negligible. a. Determine the speed of the dart just before it strikes the block. b. Calculate the horizontal distance d between the launching point of the dart and a point on the floor directly below the block. c. Calculate the speed of the block just after the dart strikes. d. Calculate the angle ? through which the dart and block on the string will rise before coming momentarily to rest. e. The block then continues to swing as a simple pendulum. Calculate the time between when the dart collides with the block and when the block first returns to its original position. f. In a second experiment, a dart with more mass is launched at the same speed and angle. The dart collides with and sticks to the same wooden block. i. Would the angle ? that the dart and block swing to increase, decrease, or stay the same? _____ Increase _____ Decrease _____ Stay the same Justify your answer. ii. Would the period of oscillation after the collision increase, decrease, or stay the same? _____ Increase _____ Decrease _____ Stay the same Justify your answer.

Transcript

00:01 Hello, in this question we are given that a small dart of mass 0 .02 kilogram is launched at an angle of 30 degree.

00:08 Okay, so it is launched at an angle of this is 30 degree with a speed of 10 meter per second.

00:17 And its mass is worth 0 .02 kilogram.

00:25 Okay, when it reaches its highest point somewhere over here, it collides with the block.

00:40 Horizontal which is the highest point of the path is moving horizontal it collides with dust and stick with the wooden block of mass 0 .1 kilogram the mass of the block is 0 .1 kilogram and this length of the string was 1 .2 meter okay it is given now the block and darts they start swimming up until the string makes an angle of theta with the vertical okay and air resistance is assumed to be upset now we need to calculate in the first part we need to calculate at with which speed it will hit the log okay at the topmost point it will only have the horizontal velocity component so here horizontal velocity component is what 10 cost 30 and vertical component is what 10 sine 30 so the horizontal component at this point would be 5 root 3 if we write it in more simplified form it would be 5 into root 3 that would be 8 .66 meter per second okay this is the first part now what does the second part says in the second part it says that we need to calculate the horizontal distance d between the launching point of the dart and the point on the floor directly below the block we need to calculate this horizontal distance so this horizontal distance can be calculated using the time required to reach up to the top would be for this block time required would be to make its vertical component of velocity 0 would be u is equal to sorry v is equal to u plus 80 so v here is zero u is 10 sine 30 it means 5 and g is minus 10 into t that must be equal to 0 so time is 0 .5 seconds so in 0 .5 seconds, how much horizontal distance it can travel is 8 .66 into 0 .5 that would be 4 .33 meter.

03:05 So for the b part, it would be answered.

03:07 For the a part, this would be the answer.

03:09 Now, c part.

03:11 What does the c part say? as per c part, we need to calculate the speed of block just after the dart strikes.

03:19 Okay, after the collision, they both combines with each other.

03:24 So linear momentum will be conserved only and both of them will move with the same final velocity.

03:31 Okay, so initially only dart is moving 0 .02 kilogram into 8 .66 that must be equal to 0 .12 into b.

03:43 So from here, v would be what? when i calculate this value, it would be now this velocity would be 1 .44 meter per second.

04:02 Okay this is what c part says now the d part as per depart it is required we need to calculate the angle theta that means that the system of dart and block will make when it will come momentarily at rest okay if this is the system if it rises by an angle theta from here it was moving with a speed of 1 .443 meter per second and here it will stop.

04:36 So what will be the change in height? change in height would be if this was some 1 .2 meter and this is 1 .2 and then this distance would be what 1 .2 cost theta.

04:52 So change in height would be 1 .2 minus 1 .2 cost theta.

04:58 So this would be increase in potential energy so all the momentum all the kinetic energy will be converted to potential energy.

05:08 So so m g h will be equal to half m v square.

05:15 So from here we get m and m both of these ms would cancel out...

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