Linear Impulse and Momentum (impact) (example continued)

Recall that impulse and momentum is a vector relation. (components in x and y directions)

Linear impulse and momentum for the bowling ball:

˗ ∫ Fdt i = M v_Bfx i ˗ M v_o i + M v_Bfy j ˗ M v_Biy j The x and y components yield:

˗ ∫ Fdt i = M v_Bfx i ˗ M v_o i (1)

and v_Bfy = v_Biy (result since there is no impulse on the bowling ball in the y-direction)

Linear impulse and momentum for the head pin: ∫ Fdt i = m v_Pfx i + m v_Pfy j

The x and y components yield: ∫ Fdt i = m v_Pfx i (2)

and v_Pfy = 0 (result since there is no impulse on the headpin in the y-direction)

Add equations (1) and (2). The result is

M v_Bfx i ˗ M v_o i + m v_Pfx i = 0 Then in scalar form:

M v_Bfx ˗ M v_o + m v_Pfx = 0 (3)

and M v_Bfy ˗ M v_Biy = 0 (4)

The unknowns are v_Bfx , v_Pfx , and v_Bfy By (4) v_Bfy = v_Biy (result)

Still there are two unknowns and only one equation (3).

Next use the coefficient of restitution to relate the normal speeds of the bowling ball and

head pin prior to and after impact. Click here to continue with this example.

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